The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 6.00×1010 atoms of zinc emitting light in the instrument flame at any given instant, what energy (in joules) must the flame continuously supply to achieve this level of emission?

Answers

Answer 1
2,405 atoms I believe

Related Questions

When pressure is increased on the following equilibrium, where will the shift be? 3H2 + N2 2NH3

Answers

Answer:

Explanation:

it is based on le chatliers principles

the left side of reaction you have 4 moles , where as at the right hand side you have 2 moles,,,,

so when you increase the pressure the reaction will shift towards the lower moles producing reaction that is reaction move towards forward in you case.

Calculate Keq for these reactions and predict if the equilibrium will lie to the right or to the left as written. (You may enter your answer in scientific notation, e.g. 1.0*10^-9. Enter your answer to two significant figures.) Reaction 1: + + pKa = 9 pKa = 38 Keq = Equilibrium position = _______ Reaction 2: + + pKa = 35 pKa = 25 Keq = Equilibrium position = _______

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

For reaction 1

    [tex]K_{eq} = 10^{29}[/tex]

     The equilibrium position is to the right

For reaction 2

       [tex]K_{eq} = 10^{-6.66}[/tex]

        The equilibrium position is to the left

Explanation:

Generally  [tex]pKa[/tex] is mathematically evaluated as  

[tex]pKa = pKa _ \ {left }} - pKa _ \ {right }}[/tex]

And equilibrium position [tex]K_a[/tex] is mathematically evaluated as [tex]K_{eq} = 10^\ {-pK_a}[/tex]

From the question we are told that

For reaction 1

         [tex]pKa_\ {left}} \ = 9[/tex]

        [tex]pKa_\ {right }} \ = 38[/tex]

So

       [tex]pKa = 9-38[/tex]

       [tex]pKa =-29[/tex]

So  [tex]K_{eq} = 10^{-(-29)}[/tex]

      [tex]K_a = 10^{29}[/tex]

This implies that the equilibrium position is to the right

   For reaction 2

       [tex]pKa_\ {left}} \ = 15.9[/tex]

       [tex]pKa_\ {right }} \ = 9.24[/tex]

So

       [tex]pKa = 15.9-9.24[/tex]

       [tex]pKa = 6.66[/tex]

So  [tex]K_{eq} = 10^{-(6.66)}[/tex]

      [tex]K_{eq} = 10^{-6.66}[/tex]

This implies that the equilibrium position is to the left

Calculate the osmotic pressure of a solution prepared by dissolving 65.0 g of Na2SO4 in enough water to make 500 mL of solution at 20°C. (Assume no ion pairing – in other words, assume that the electrolyte completely dissociates into its constituent ions.)

Answers

Answer:

66.0 atm

Explanation:

We can calculate the osmotic pressure (π) using the following expression.

[tex]\pi = i \times M \times R \times T[/tex]

where,

i: van 't Hoff indexM: molarityR: ideal gas constantT: absolute temperature

Step 1: Calculate i

Sodium sulfate completely dissociates according to the following equation.

Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻

Since it produces 3 ions, i = 3.

Step 2: Calculate M

We can calculate the molarity of Na₂SO₄ using the following expression.

[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]

Step 3: Calculate T

We will use the following expression.

K = °C + 273.15

K = 20°C + 273.15 = 293 K

Step 4: Calculate π

[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]

Consider each pair of compounds listed below and determine whether a fractional distillation would be necessary to separate them or if a simple distillation would be sufficient.

a. Ethyl acetate and hexane
b. Diethyl Ether and 1-butanol
c. Bromobenzene and 1,2-dibromobenzene

Answers

It’s b the answer Distillationis a process of separation of liquids having significantly different boiling points.SimpleDistillationis used if the components have widely different

When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation

Answers

All done for you no worries

When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. This reaction describes a non-Markovnikov orientation.

In the reaction between an unsymmetrical alkene (such as propene) and N-bromosuccinimide (NBS) in the presence of aqueous dimethyl sulfoxide (DMSO), the major product is formed with the bromine atom bonded to the less highly substituted carbon atom of the alkene.

In Markovnikov's addition, the major product is formed by adding the electrophile (in this case, the bromine atom) to the carbon atom with more hydrogen atoms bonded to it. However, the given reaction exhibits non-Markovnikov selectivity, as the bromine atom adds to the less substituted carbon atom.

This non-Markovnikov selectivity can be attributed to the presence of DMSO, which acts as a polar solvent and helps generate a bromine radical (Br•). The radical intermediate can then undergo reaction with the alkene, leading to the observed regioselectivity where the bromine atom adds to the less substituted carbon. This process is known as a radical addition reaction.

Hence, the reaction demonstrates a non-Markovnikov orientation due to the addition of the bromine atom to the less highly substituted carbon atom of the propene molecule.

Learn more about the Markovnikov rule here:

https://brainly.com/question/33423745

#SPJ 2

A compound is known to be Na2CO3, Na2SO4, NaOH, NaCl, NaC2H3O2, or NaNO3. When a barium nitrate solution is added to a solution containing the unknown a white precipitate forms. No precipitate is observed when a magnesium nitrate solution is added to a solution containing the unknown. What is the identity of the unknown compound

Answers

Answer:

Na₂SO₄

Explanation:

Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.

Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:

Na₂SO₄

A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0mM solution of M(NO3)3 and the right half cell with a 3.0M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 C.

Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode.

Answers

Answer:

1.The electrode on the right is positive

2. 0.058V

Explanation:

The above cell is a concentration cell.

A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.

In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).

Part 2: Please, see the attachment below for the calculations.

Precision can be defined as the?

Answers

Answer:Precision can be defined as the. reproducibility of a measured value. Precision is how close the measured values are to each others. In contrast with accuracy, accuracy is the agreement between a measured value and an accepted value.

Explanation:

Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . If of sodium bromide is produced from the reaction of of hydrobromic acid and of sodium hydroxide, calculate the percent yield of sodium bromide.

Answers

Answer:

The percentage yield is 50%

An ideal gaseous reaction occurs at a constant pressure of 35.0 atm and releases 66.8 kJ of heat. Before the reaction, the volume of the system was 8.20 L. After the reaction, the volume of the system was 2.21 L. Calculate the total change in internal energy for the system. Enter your answer numerically in units of kJ.

Answers

Answer:

U = -45.557kj

Explanation:

Before we can calculate the totally internal energy change in kilojoules firstly we need to calculate W

U=q + w .

We know that

w = PΔ V

where P is the pressure of

and V is the volume

then we can calculate the work

w = 35 atm * ( 8.20L - 2.21L)

W=35atm* 5.99L

W=209.65atmJ

But 1 atm = 101.325J

then ,

w = 209.65* 101.325 J = 21242.79 J

let us convert it to Kj

But we know that 1kJ = 10^3 J .

Then w = 21.243 kJ .

Then we can now calculate the internal energy as

U = 21.243- 66.8 kJ = -45.557kj

But we know that heat was released. Theeefore, the total internal energy change was -45.557kj

A stock solution of HNO3 is prepared and found to contain 14.9 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution

Answers

Answer:

[tex]0.745~M[/tex]

Explanation:

In this case, we have a dilution problem. So, we have to use the dilution equation:

[tex]C_1*V_1=C_2*V_2[/tex]

Now, we have to identify the variables:

[tex]C_1~=~14.9~M[/tex]

[tex]V_1~=~25~mL[/tex]

[tex]C_2~=~?[/tex]

[tex]V_2~=~0.5~L[/tex]

Now, we have different units for the volume, so we have to do the conversion:

[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]

Now we can plug the values into the equation:

[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]

I hope it helps!

The temperature program for a separation starts at a temperature of 50 °C and ramps the temperature up to 270 °C at a rate of 10 °C/minute. Which statement is NOT true for this separation?
A) At 10 °C/minute, a total of 22 minutes is needed to reach 270 oC.
B) Strongly retained solutes will remain at the head of the column while the temperature is low.
C) Weakly retained solutes will separate and elute early in the separation.
D) The vapor pressure of strongly retained solutes will increase as temperature increases.
E) Strongly retained analytes will give broad peaks.

Answers

Answer:

The correct answer to the question is Option E (Strongly retained analytes will give broad peaks).

Explanation:

The other options are true because:

A. Initial temp = 50 °C

   Final temp =  270 °C

Differences in temp = 270 - 50 = 220°C

Rate =  10 °C/minute.

So, at 10 °C/minute,

total of 220°C /10 °C = number of minutes required to reach the final temp.

220/10 = 22 minutes

B. A column has a minimum and maximum use temperature. Solutes that are already retained would remain stationary while temperatures are low. This would only change if there is an increase in temperature. Heat transfers more energy to the liquid which would make the solute interact with the column phase.

C. Weakly retained solutes may contain larger molecules, will separate by absorbing into the solvent early in separation making the mobile phase separates out into its components on the stationary phase.

D. Retained solute's vapor pressure is higher at higher temperatures making it possible for particle to escape more from the solute when the temperature is high than when it is low.

The boiling point of diethyl ether, CH3CH2OCH2CH3, is 34.500 °C at 1 atmosphere. Kb(diethyl ether) = 2.02 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.94 grams of the compound were dissolved in 279.5 grams of diethyl ether, the solution began to boil at 35.100 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?

Answers

Answer:

The correct answer is 179.94 g/mol.

Explanation:

Based on the given question, the boiling point of diethyl ether us 34.500 degree C at 1 atm pressure. The boiling point of the solution given is 35.100 degree C. The Kb of diethyl ether given is 2.02 degree C/m. The weight of the compound given is 14.94 grams, the weight of the solvent (diethyl ether) is 279.5 grams.

The molecular weight of the compound can be determined by using the formula,

deltaTb = Kb * molality

Tb-To = Kb * molality

Tb-To = Kb*wt/mol.wt*1000/w (solvent)

35.100 - 34.500 = 2.02 * 14.94 / mol. wt * 1000 g / 279.5 g

0.6 = 2.02 * 53.45/ mol.wt

mol. wt = 2.02*53.45/0.6

mol. wt = 179.94 g/mol

Hence, the molecular weight of the compound is 179.94 gram per mol.

Problem PageQuestion Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 60. g of hexane is mixed with 74.5 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.

Answers

Answer:

43.45g of water would be produced from the reaction.

Explanation:

Liquid became reacts with oxygen to produce carbon dioxide and water.

This type of reaction is known as combustion reaction between alkanes.

Equation of reaction.

Assuming the reaction occurs in an unlimited supply of oxygen,

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

From the above equation of reaction,

2 moles of C₆H₁₄ reacts with 19 moles of O₂ to produce 14 moles of H₂O.

To find the theoretical mass,

Number of moles = mass / molar mass

Molar mass of C₆H₁₄ = 86g/mol

Molar mass of O₂ = 16g/mol × 2 = 32g/mol

Molar mass of H₂O = 18g/mol

Mass of H₂O = number of moles × molar mass

Mass of H₂O = 14 × 18 = 252g

Mass of C₆H₁₄ = number of moles × molar mass

Mass of C₆H₁₄ = 2 × 86 = 172g

Mass of O₂ = number of moles × molar mass

Mass of O₂ = 19 × 32 = 608g

From the equation of reaction,

172g of C₆H₁₄ reacts with 608g of O₂ to produce 252g of H₂O

(172 + 608)g of reactants produce 252g of H₂O

780g of reactants produce 252g of H₂O

(60 + 75.5)g of reactants will produce a x g of H₂O

780g of reactants = 252g of H₂O

134.5g of reactants = x g of H₂O

X = (134.5 × 252) / 780

X = 43.45g of H₂O

Therefore, 43.45g of H₂O would be produced from 60g of hexane and 74.5g of oxygen

Answer:

[tex]m_{H_2O}=30.9gH_2O[/tex]

Explanation:

Hello,

In this case, the combustion of hexane is given by:

[tex]C_6H_{14}+\frac{19}{2} O_2\rightarrow 6CO_2+7H_2O[/tex]

The next step is to compute the reacting moles of hexane:

[tex]n_{C_6H_{14}}=60gC_6H_{14}*\frac{1molC_6H_{14}}{86gC_6H_{14}} =0.698molC_6H_{14}[/tex]

Then, the moles of hexane consumed by 74.5 g of oxygen using the molar ratio in the chemical reaction (1:19/2):

[tex]n_{C_6H_{14}}=74.5gO_2*\frac{1molO_2}{32gO_2} *\frac{1molC_6H_{14}}{19/2molO_2} =0.245molC_6H_{14}[/tex]

Therefore, as less moles of hexane are consumed by oxygen, it is in excess, so we compute the mass of water produced by the consumed 0.245 moles of hexane:

[tex]m_{H_2O}=0.245molC_6H_{14}*\frac{7molH_2O}{1molC_6H_{14}}*\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=30.9gH_2O[/tex]

Best regards.

uses of sodium chloride in daily life​

Answers

Answer:

sodium chloride can be used as salt

extraction sodium metal by electrolysis

a common chemical in laboratory experiments

Answer:

sodium chloride can be used as preservatives,

in preserving foods.

Phosphofructokinase is a four‑subunit protein with four active sites. Phosphofructokinase catalyzes step 3 of glycolysis, converting fructose‑6‑phosphate to fructose‑1,6‑bisphosphate. Phosphoenolpyruvate (PEP) is the product of step 9 of glycolysis. The PEP concentration in the cell affects phosphofructokinase activity.Select the true statements about PEP regulation of phosphofructokinase.
1. PEP is a feedback inhibitor of phosphofructokinase.
2. The apparent affinity of phosphofructokinase for its substrate increases when PEP binds.
3. PEP is a positive effector of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
5. PEP competes with fructose-6-phosphate for the active site of phosphofructokinase.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.

Answers

Answer:

1. PEP is a feedback inhibitor of phosphofructokinase.

4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.

6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.

Explanation:

Phosphofructokinase-1, PFK-1, is an allosteric enzymes composed of four protein subunits.

Allosteric enzymes are enzymes that function through non-covalent binding of allosteric modulators which may be activators or inhibitors. They produce a characteristic velocity versus substrate sigmoidal curve. PFK-1 has a separate binding site for its substrate, fructose-6-phosphate and it's allosteric modulators: ATP, ADP or phosphoenolpyruvate, PEP.

The enzyme can exist in two conformations, the T-state (tense) or the R-state (resting). Binding of substrate causes a conformational change from T-state to R-state, whereas binding of allosteric inhibitors returns it to the T-state.

PEP, the product of step 9 in glycolysis, is an allosteric inhibitor of PFK-1. When it binds to the the allosteric site, it leads to conformational changes in PFK-1 from the R-state to the T-state which reduces the enzymes ability to bind the substrate. These changes are responsible for the sigmoidal velocity/substrate curve in allosteric enzymes.

Therefore, the true statements from the options above are 1, 4, 6.

Options 2,3 and 5 are wrong because PEP is a negative effector of PFK-1, thus its binding reduces the affinity of PFK-1 for its substrate. Also, PFK-1 being an allosteric enzyme has separate binding sites for its substrate and its modulators. Thus, there is no competition for active site binding by substrate and modulators.

Which process is a physical change

Answers

Answer:

a physical change is something that has not been modified chemically and can possibly be changed back to the state it was once before. A physical change keeps all the same atoms and none of them is modified.

Example:

When a block of clay is morphed into a giraffe statue, it can be morphed back to its original state. If someone burnt the block of clay, the atoms would be modified and it would be unable to go back to its previous state.

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(If you're referring to a question with these answers)

A. iron rusting

B. milk turning to curd

C. water boiling

D. paper burning

E. hard water staining pipes

-----------------------------------------------------------------------------------------------------------------

Answer:

C. Water Boiling

(If you are referring to a question with these answers I think this is the correct answer if not I do apologize)

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Calculate the pH of this solution 0.0043 M of H2SO4=

Answers

Answer:

pH = - log [concentration]

pH = - log (0.0043M)

pH = 2.37

Given that S is the central atom, draw a Lewis structure of OSF4 in which the formal charges of all atoms are zero. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.

Answers

Answer:

Here's what I get  

Explanation:

A Lewis structure shows the valence electrons surrounding the atoms.

Your structure has two problems:

It shows too many valence electrons It violates the octet rule for O — there are 10 electrons around the O atom.

Here's one way to draw a Lewis structure.

1. Draw a trial structure

Make F and O terminal atoms and give each one an octet (Fig. 1).

2. Count the valence electrons in the trial structure

5 BP + 15 LP  = 10 + 30 = 40 electrons

3. Check the number of valence electrons available  

1 S =   1 × 6 =  6 electrons

1 O =   1 × 6 =   6

4 F  = 4 × 7 = 28                    

     TOTAL = 40 electrons

The trial structure has the correct number of electrons.

4. Determine the formal charge on each atom.

To get the formal charges, we cut the covalent bonds in half.

Each atom gets the electrons on its side of the cut.

Formal charge = valence electrons in isolated atom - electrons on bonded atom

FC = VE - BE  

(a) On S

VE = 6

BE = 5 bonding electrons = 5

FC = 6 - 5 = +1

(b) On O:

VE = 6

BE = 3 LP(six electrons) + 1 bonding electron  = 7

FC = 6 - 7 = -1

(c) On F:

VE = 6

BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7

FC = 7 - 7 = 0

5. Minimize the formal charges

We must rearrange the valence electrons so that S gets one more and O gets one fewer.

Move a lone pair from the O to make an S=O double bond (Fig. 2).

6. Recalculate the formal charges

(a) On S

VE = 6

BE =  (3 bonding electrons) = 6

FC = 6 - 6 = 0

(b) On O:

VE = 6

BE = 2 LP(four electrons) + 2 bonding electrons = 6  

FC = 6 - 6 = 0

Fig. 2 shows the Lewis structure in which all atoms have a formal charge of  zero.

The formal charge of the atoms can be concluded zero with the bond formation between the sulfur and oxygen atom.

The lewis structure can be defined as the dot structure of the valence bond with the bonded atoms. The formal charge can be calculated with the difference in the valence electrons and the bonding electrons.

The formal charge of an atom can be zero when the valence electrons and the bonding electrons are equal. In the structure of [tex]\rm OSF_4[/tex], the formal charge has been assigned zero with the bond formation resulting in the valence electrons and bonding electrons being equal.

The lewis structure with the central S atom has been attached.

For more information about lewis structure, refer to the link:

https://brainly.com/question/4144781

I WILL GIVE BRAINLIEST

Answers

Molarity= no. of molecules of solute /1 liter
one moles of sodium hydroxide =49 gm of sodium hydroxide
So we can say that if we want to prepare 1 molar NaOH solution then we need 40 gm NaOH dissolve in one liter of water so it can become one 1 molar NaOH solution.

Dry chemical hand warmers utilize the oxidation of iron to form iron oxide according to the following reaction: 4Fe(s)+3O2(g)→2Fe2O3(s) Standard thermodynamic quantities for selected substances at 25 ∘C Reactant or product ΔH∘f(kJ/mol) Fe(s) 0.0 O2(g) 0.0 Fe2O3(s) −824.2 Calculate ΔH∘rxn for this reaction.

Answers

Answer:

-1648.4 kJ/mol

Explanation:

Based on Hess's law:

ΔHr = ∑n×ΔH°f(products) - ∑n×ΔH°f(reactants)

In the reaction:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

ΔHr = 2 ΔH°f {Fe₂O₃} - (4ΔH°f {Fe(s)} + 3ΔH°f{O₂(g)}

As:

ΔH°f {Fe₂O₃} = -824.2kJ/mol

ΔH°f {Fe(s)} = 0.0kJ/mol

ΔH°f{O₂(g)} = 0.0kJ/mol.

Thus,

ΔHr = 2 ₓ -824.2kJ/mol =

-1648.4 kJ/mol

Answer:

-1648.4 kJ

Explanation:

The product has the only nonzero heat of formation, so it is the only value needed to calculate the enthalpy of this reaction. Normally, you would want to express the enthalpy of a reaction with respect to one mole of a chemical species, whether it is a reactant or product. However, since the balanced chemical equation contains only coefficients greater than 1, you should consider how the enthalpy relates to one mole of each substance according to the coefficients. In other words,  − 1648.4  kJ  of heat is released when 4  mol  of  Fe  reacts with 3  mol  of  O2  to produce 2  mol  of  Fe2O3 .

how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure

Answers

Answer:

1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.

Explanation:

If 200.4g of water is mixed with 101.42g of salt the mass of the final solution would be reported as

Answers

Answer:

301.8 g

Explanation:

We prepare a solution with 200.4 g of water (solvent) and 101.42 g of salt (solute). The mass of the solution is equal to the sum of the mass of the solvent and the mass of the solute.

m(solution) = m(solute) + m(solvent)

m(solution) = 200.4 g + 101.42 g

m(solution) = 301.8 g (we round-off to one decimal according to the significant figures rules)

How many grams of the salt CaF2 (g) are formed when 15.7 mL of 0.612 M KF reacts with an excess of aqueous calcium bicarbonate (Ca(HCO3)2) via a metathesis reaction?

Answers

Answer:

[tex]m_{CaF_2}0.375gCaF_2[/tex]

Explanation:

Hello,

In this case, for the studied reaction:

[tex]2KF+Ca(HCO_3)_2\rightarrow CaF_2+2KHCO_3[/tex]

Thus, the first step is to compute the reacting moles of potassium fluoride by using its volume and molarity:

[tex]n_{KF}=0.0157L*0.612\frac{mol}{L} =9.61x10^{-3}molKF[/tex]

Then, we apply the 2:1 molar ratio between potassium fluoride and calcium fluoride to compute the produced moles of calcium fluoride:

[tex]n_{CaF_2}=9.61x10^{-3}molKF*\frac{1molCaF_2}{2molKF} =4.80x10^{-3}molCaF_2[/tex]

Finally, by using the molar mass of calcium fluoride (78.07 g/mol) we can compute its produced grams:

[tex]m_{CaF_2}=4.80x10^{-3}molCaF_2*\frac{78.07gCaF_2}{1molCaF_2} \\\\m_{CaF_2}0.375gCaF_2[/tex]

Best regards.

What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together

Answers

Answer:

B. Mixing a solute and a solvent

Explanation:

Hello,

In this case, solutions are defined as liquid homogeneous mixtures formed when two substances having affinity are mixed. It is important to notice that the two substances are known as solute, which is added to other substance that is the solvent. Therefore, answer is B. Mixing a solute and a solvent.

Notice that when two insoluble substances are mixed no solution is formed. Furthermore, if two solutes together or a solute and a precipitate are mixed, no liquid homogeneous solution is formed, as commonly solutes are solid, nevertheless, when liquid, one should have to act as the solvent.

Best regards.

Answer:

B. Mixing a solute and a solvent

Explanation:

ap3x

How many moles of solute are contained in the following solution: 15.25 mL of a 2.10 M CaCl₂

Answers

Answer:

0.032moles

Explanation:

2.10moles in 1000ml what about 15.25ml

(15.25×2.10)÷1000

0.032moles

You are trying to recrystallize compound X. You consider using ethyl acetate as your recrystallizing solvent and test a small amount of compound X with ethyl acetate. You find that compound X is soluble in ethyl acetate at room temperature and at boiling. Is ethyl acetate a good recrystallization solvent? No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath. Yes, you want the sample to fully dissolve at room temperature and boiling so that it will crystallize in the ice bath. Yes, you can only be sure that all the impurities dissolved if the sample is soluble at room temperature

Answers

Answer:

No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath.

Explanation:

For a solvent to be adequate it must completely dissolve the substance to be purified when it is hot, that is, at boiling temperature only. It should be practically insoluble when the solvent is cold or at room temperature. This must occur in this way since impurities must be removed by hot filtering or dissolved in the mother liquor.

A 75 gram solid cube of mercury (II) oxide has a density of 2.4 x 103 kg/m3 .
What is the length of one side of the cube in cm?

The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?

The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?

Answers

Answer:

0.031 m

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

Chemical change

Element

Explanation:

A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³.  What is the length of one side of the cube in cm?

Step 1: Convert the mass to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]

Step 2: Calculate the volume (V) of the cube

[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]

Step 3: Calculate the length (l) of one side of the cube

We will use the following expression.

[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]

The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?

The balanced chemical equation is:

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

This is a chemical change because new substances are formed.

The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?

The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.

Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4.

Answers

Answer:

4.41

Explanation:

Step 1: Write the balanced equation

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

Step 2: Calculate the respective concentrations

[tex][CO]_i = \frac{0.500mol}{5.00L} = 0.100M[/tex]

[tex][H_2]_i = \frac{1.500mol}{5.00L} = 0.300M[/tex]

[tex][H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M[/tex]

Step 3: Make an ICE chart

        CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

I       0.100      0.300        0            0

C         -x           -3x          +x          +x

E    0.100-x    0.300-3x     x            x

Step 4: Find the value of x

Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396

Step 5: Find the concentrations at equilibrium

[CO] = 0.100-x = 0.100-0.0396 = 0.060 M

[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M

[CH₄] = x = 0.0396 M

[H₂O] = x = 0.0396 M

Step 6: Calculate the equilibrium constant (Kc)

[tex]Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41[/tex]

Which of the following best describes isotopes?


An element with the same number of neutrons, but a different number of protons.


An element with the same number of protons, but a different number of electrons.


An element with the same number of electrons, but a different number of neutrons


An element with the same number of protons, but a different number of neutrons

Answers

Answer: An element with the same number of protons, but a different number of neutrons

Explanation:

The # of protons in an atom is what determines what atom it is (hydrogen has 1 proton, helium has 2 protons, etc ...). You cannot change the number of protons in an atom without changing what element the atom is.

The number of electrons in atoms varies greatly because electrons are constantly gained, lost, and shared during chemical reactions.

An isotope is a variation of the same element (so they must have the same # of protons) that have different masses (and therefore a different number of neutrons).

The answer is the fourth choice, "An element with the same number of protons, but a different number of neutrons"

The isotopes refer to an element that consists of a similar number of protons but have a distinct no of neutrons.

What are isotopes:

It is considered to be the members of the family with respect to the elements that consist of a similar number of protons but have a distinct no of neutrons. The no of protons in the nucleus measured the atomic number of elements based on the periodic table.

Therefore, the fourth option is correct.

Learn more about isotopes here: https://brainly.com/question/22318349

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