Answer:
0.136g
Explanation:
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
[tex]Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)[/tex]
Initial mole of Co(NO3)2 [tex]=\frac{mass}{molar mass}[/tex]
[tex]=\frac{5.00}{182.94} \\\\=0.02733mol[/tex]
Mole of Co(NO3)2 in final solution
[tex]=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol[/tex]
Mole of NO3- in final solution = 2 x Mole of Co(NO3)2
[tex]=2\times 0.001093\\\\=0.002186mol[/tex]
Mass of NO3- in final solution is mole x Molar mass of NO3
[tex]=0.002186\times62.01\\\\=0.136g[/tex]
The final solution contains 0.24 g of nitrate ion.
Number of moles of Co(NO3)2 = 5.00 g/183 g/mol = 0.027 moles
Number of moles = concentration × volume
concentration = Number of moles /volume
Volume of solution = 100 mL or 0.1 L
concentration = 0.027 moles/0.1 L = 0.27 M
Using the dilution formula;
C1V1 = C2V2
C1 = 0.27 M
V1 = 4.00 mL
C2 = ?
V2 = 275. mL
C2 = C1V1/V2
C2 = 0.27 × 4.00/ 275
C2 = 0.0039 M
Number of moles of NO3- ion in Co(NO3)2 = 0.0039 M × 62 g/mol = 0.24 g
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Each of the insoluble salts below are put into 0.10 M hydrobromic acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution ?
a. Calcium sulfite
b. Calcium fluoride
c. Silver bromide
Answer:
A. Solubility of calcium sulfite increases
B. Solubility of calcium fluoride increases
C. Solubility of Silver bromide decreases
Explanation:
The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.
Case 1. Calcium sulfite
The dissociation reaction of CaSO3 is given below:
CaSO3 ----> Ca²+ + SO3²-
SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;
H+ + SO3²- ---> HSO3-
This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.
Case 2. Calcium fluoride
The dissociation reaction of calcium fluoride (CaF2) is shown below.
CaF2 ----> Ca²+ + 2F-
Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:
H+ + F- ---->. HF
Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.
Case 3: Silver bromide
The dissociation reaction of AgBr is as follows:
AgBr ----> Ag+ + Br-
The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.
The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.
Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.
Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.
However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.
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A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.
Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
Explanation:
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]
[tex]k_b[/tex] = boiling point constant = ?
m = molality
[tex]w_2[/tex] = mass of solute (urea) = 55.4 g
[tex]w_1[/tex] = mass of solvent X = 500 g
[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]
[tex]k_b=2.4^0C/m[/tex]
Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]
How many grams of 02 are needed to react with 7.50g of ethanol
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO 3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH° rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water.
Answer:
THE STANDARD HEAT OF NEUTRALIZATION OF THE BASE SODIUM HYDROXIDE BY THE ACID HYDROGEN TRIOXONITRATE V ACID IS -56 kJ / mol.
Explanation:
Volume of 0.3 M NaOh = 100 mL
Volume of 0.3 M HNO3 = 100 mL
Initail temp of NaOH and HNO3 = 35 °C = 35 + 273 K = 308 K
Final temp. of mixture = 37 °C = 37 + 273 K = 310 K
We can make the following assumptions form the question given:
1. specific heat of the reaction mixture is the same as the specific heat of water = 4.2 J/g K
2. the toal mass of the reaction mixture is 200 mL = 200 g since no heat is lost to the calorimeter or surrounding.
3. initail temperature of the reaction mixture is equal to the average temperature of the two reactant solutions
= ( 308 + 308 /2) = 308 K
4. Rise in temeperature for the reaction = 310 -308 K = 2 K
Then the total heat evolved during the reaction = mass * specifc heat capacity * temperature change
Heat = 200 g * 4.2 J/g K * 2 K
Heat = 1680 J
EQUATION FOR THE REACTION
HNO3 + NaOH -------> NaNO3 + H20
From the equation, 1 mole of HNO3 reacts with 1 mole of NaOH to prouce mole of water.
100 mL of 0.5 M HNO3 contains 100 * 0.3 /1000 = 0.03 mole of acid
This result is same for the base NaOH = 0.03 mole of base
So therefore,
0.03 mole of acid will react with 0.03 mole of base to produce 0.03 mole of water to evolved 1680 J of heat energy.
The production of 1 mole of water will evolve 1680 / 0.03 J of heat
= 56 000 J or 56 kJ of heat energy per mole of water.
So therefore, 1the standard heat of neutralization of sodium hydroxide by trioxoxnitrate V acid is -56 kJ/mol.
What kind of solid is crystalline boron (B)?
A. lonic solid
B. Metallic solid
C. Molecular solid
D. Network solid
Answer:
D
Explanation:
gr. 2.3 at 25°C; valence +3. Boron is a nonmetallic element existing as a dark brown to black amorphous powder or as an extremely hard, usually jet-black to silver-gray, brittle, lustrous, metallike crystalline solid
it is a network solid, a lattice of many covalent bonds (like diamond, except that it is black rather than transparent).
Network solid kind of solid is crystalline boron (B). Hence, option D is correct.
What is Network solid?A network solid is a solid where all the atoms are covalently bonded in a continuous network.
Boron is a nonmetallic element existing as a dark brown to black amorphous powder or as an extremely hard, usually jet-black to silver-grey, brittle, lustrous, metallike crystalline solid
It is a network solid, a lattice of many covalent bonds (like a diamond, except that it is black rather than transparent).
Hence, option D is correct.
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ilicon has three naturally occurring isotopes (Si-28, Si-29, and Si-30). Masses and natural abundances for two isotopes are listed here. Isotope Mass (amu) Abundance (%) Si-28 27.9769 92.2 Si-29 28.9765 4.67 Si-30 ? ? Part A Find the natural abundance of Si-30 . Express your answer using two significant figures.
Answer:
Part A
The natural abundance of Si-30 = 3.1
Part B
The mass of Si-30 = 29.9551 amu
Explanation:
Part A
The sum of the natural abundances = 100
Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%.
Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67%
Si-30's mass and natural abundance are unknown.
Natural abundance for Si-30 = 100% - 92.2% - 4.67% = 3.13% = 3.1% to 2 s.f.
Part B
The total atomic mass of an element is an addition combination of the mass and natural abundances of all the isotopes of that element.
Molar mass of Silicon normally = 28.0855 amu
Let the mass of Si-30 be m
28.0855 = (27.9769×0.922) + (28.9765×0.0467) + (m×0.0313)
28.0855 = 27.14790435 + 0.0313m
0.0313m = 28.0855 - 27.14790435 = 0.93759565
m = (0.93759565/0.0313) = 29.9551325879 amu = 29.9551 amu
Hope this Helps!!
Part A: The natural abundance of Si-30 = 3.1
Lets solve the question:
The sum of the natural abundances = 100 Si-28 has a mass of 27.9769 amu and a natural abundance of 92.2%. Si-29 has a mass of 28.9765 amu and a natural abundance of 4.67% Si-30's mass and natural abundance are unknown.Natural abundance (NA) refers to the abundance of isotopes of a chemical element as naturally found on a planet.
Natural abundance for Si-30 = [tex]100\% - 92.2\% - 4.67\% = 3.13\% = 3.1\%[/tex] in significant figures.
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what is the sign of Mercury
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
Constructive interference occurs when the compression of one wave meets
up with the compression of a second wave.
A. True
B. False
Answer:
True
Explanation:
Coefficient of balanced equation: __Fe + ___020) — _Fe_036)
Answer:
- Four for iron, three for oxygen and 2 for iron (III) oxide:
[tex]4Fe+3O_2\rightarrow 2Fe_2O_3[/tex]
Explanation:
Hello,
In this case, the oxidation of iron is a widely acknowledged reaction occurring in ships and other machines exposed to the air or highly oxidizing medias. Thus, by the effect of oxygen, iron undergoes oxidation typically to iron (III) oxide:
[tex]Fe+O_2\rightarrow Fe_2O_3[/tex]
Nonetheless, the law of conservation of mass must be respected, therefore the coefficients balancing the reaction are four for iron, three for oxygen and 2 for iron (III) oxide:
[tex]4Fe+3O_2\rightarrow 2Fe_2O_3[/tex]
Best regards.
Indicate whether the following represents a Chemical or Physical change: Milk sours
Answer:
Chemical Change
Explanation:
Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.
The substances nitrogen monoxide and hydrogen gas react to form nitrogen gas and water. Unbalanced equation: NO (g) + H2 (g) N2 (g) + H2O (l) In one reaction, 76.2 g of H2O is produced. What amount (in mol) of H2 was consumed? What mass (in grams) of N2 is produced?
Answer:
H2 consumed 4.22 mol
N2 produced 59.107 g
Explanation:
Balanced equation:
2NO (g) + 2H2 (g) N2 (g) + 2H2O (l)
To perform the calculations, the molecular weights of the following compounds must be known:H2O MW = 18.02 g/mol
N2 MW = 28.01 g/mol
To determine the moles of H2O produced, the following formula should be used:
[tex]MW=\frac{mass}{mol}[/tex]
The value of moles is cleared:
[tex]mol=\frac{mass}{MW} =\frac{76.2g}{18.02\frac{g}{mol} } =4.22 mol[/tex]
Now, to calculate the grams of N2 consumed, we look at the balanced equation and note that 2 moles of H2 produce 1 mole of N2. Therefore, through said observation, the amount of moles of H2 consumed can be determined.2 mol H2 ⇒ 1 mol N2
4.22 mol H2 ⇒ X
[tex]X=\frac{4.22mol*1 mol}{2 mol} =2.11 mol[/tex]
To calculate the mass of H2 consumed, the molecular weight equation is used again:
[tex]mass=MW*mol=28.013\frac{g}{mol}*2.11mol=59.107g[/tex]
Asbestosis is a lung disease caused by inhaling asbestos fibers. The US Department of Health and Human Services considers a particular form of asbestos to be a carcinogen. The composition of this form of asbestos is 26.31% Mg, 20.20% Is, 1.45% H and the rest of the mass is due to oxygen. The molar mass of the compound is 277 g/mol. What is the molecular formula for the carcinogenic form of asbestos
Answer: The molecular formula for the carcinogenic form of asbestos [tex]Mg_3Si_2H_4O_9[/tex]
Explanation:
a) If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of Mg = 26.31 g
Mass of Si= 20.20 g
Mass of H= 1.45 g
Mass of O= (100-(26.31+ 20.20+ 1.45)) = 52.04 g
Step 1 : convert given masses into moles
Moles of Mg=[tex]\frac{\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac{26.31g}{24g/mole}=1.10moles[/tex]
Moles of Si=[tex]\frac{\text{ given mass of Si}}{\text{ molar mass of Si}}= \frac{20.20g}{28g/mole}=0.72moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.45g}{1g/mole}=1.45moles[/tex]
Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{52.04g}{16g/mole}=3.25moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Mg = [tex]\frac{1.10}{0.72}=1.5[/tex]
For Si =[tex]\frac{0.72}{0.72}=1[/tex]
For H=[tex]\frac{1.45}{0.72}=2[/tex]
For O =[tex]\frac{3.25}{0.72}=4.5[/tex]
The ratio of Mg : Si: H: O = 1.5 : 1 : 2 : 4.5
Converting them into whole numbers :
The ratio of Mg : Si: H: O = 3 : 2 : 4 : 9
Hence the empirical formula is [tex]Mg_3Si_2H_4O_9[/tex]
Empirical mass =[tex]3\times 24+2\times 28+4\times 1+9\times 16=276g[/tex]
Molecular mass = 277 g
[tex]n= \frac{\text {Molecular mass}}{\text {Empirical mass}}=\frac{277}{276}=1[/tex]
Thus molecular formula =[tex]1\times Mg_3Si_2H_4O_9=Mg_3Si_2H_4O_9[/tex]
How to treat stream water for drinking
Answer:
Explanation:
Boiling.
Use water filter
Use Ultraviolet Light.
Use chlorine drops
I would recomade boiling as the main
because its the easiest and cheapest Or water filter if you have one
Which of the following is evidence for a physical change? A) burning B) fizzing C) evaporating D) rusting
Answer:c
Explanation: rusting, burning and fuzzing are all examples of chemical reactions/changes.
Calcium carbonate can break down to form calcium oxide and carbon dioxide.
CaCO3 → CaO + CO2
The oxygen atoms on the product side are
Answer:
balanced because the total number of oxygen atoms is 3.
Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of the visible spectrum. Suppose a particular cone cell absorbs light with a wavelength of 434.nm. Calculate the frequency of this light. Round your answer to 3 significant digits.
Answer:
6.91 × 10¹⁴ s⁻¹
Explanation:
Step 1: Given data
Wavelength of the radiation absorbed by the cone (λ): 434 nm
Step 2: Convert the wavelength to meters
We will use the relationship 1 m = 10⁹ nm.
[tex]434nm \times \frac{1m}{10^{9}nm } =4.34 \times 10^{-7} m[/tex]
Step 3: Calculate the frequency (ν) of the radiation
We will use the following expression.
[tex]c = \lambda \times \nu[/tex]
where,
c is the speed of light (3.00 × 10⁸ m/s)
[tex]c = \lambda \times \nu\\\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}m/s }{4.34 \times 10^{-7}m }= 6.91 \times 10^{14} s^{-1}[/tex]
A stock solution will be prepared by mixing the following chemicals together:
3.0 mL of 0.00200 M KSCN
10.0 mL of 0.200 M Fe(NO3)3
17.0 mL of 0.5 M HNO3
Determine the molar concentration of Fe(NO3)3 in the stock solution.
Answer:
0.067M Fe(NO3)3
Explanation:
A stock solution is a concentrated solution that is diluted to prepare the solutions that you will use.
The volume of the stock solution is 3.0mL + 10.0mL + 17.0mL= 30.0mL.
The ratio between volume of the aliquot (10.0mL) and total volume (30.0mL) is called dilution factor, that is: 30.0mL / 10.0mL = 3
That means the Fe(NO3)3 is diluted 3 times. That means the molar concentration of the stock solution is:
0.200M / 3 =
0.067M Fe(NO3)3Which process is used to make lime (calcium oxide) from limestone (calcium carbonate)?
Answer:
Explanation:
Calcium oxide is fromed by the decomopostion of CaCO3 at high temperature.
CaCO3 ------> CaO +CO2
Hope this helps you
The core of the pyruvate dehydrogenase complex is made up of eight catalytic________that make up the_______component.
a. monomers; E1b. dimers; E2c. dimers; E3d. trimers; E2
Answer:
(D.)
The core of the pyruvate dehydrogenase complex is made up of eight catalytic trimers that make up the E2 component.
Explanation:
Eight trimers assemble as a hollow truncated cube, which forms the core of the multi-enzyme complex, known as the E2 complex in human pyruvate dehydrogenase complex.
The acetate ion is the conjugate base of the weak acid acetic acid. The value of Kb for CH3COO-, is 5.56×10-10. Write the equation for the reaction that goes with this equilibrium constant.
Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
Explanation:
[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]
Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]
The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
[tex]K_a[/tex] for [tex]CH_3COOH[/tex] :
[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]
[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]
[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
2 Points
What is the voltage of an electrolytic cell with copper and magnesium
electrodes?
A. -2.71 v
B. 2.71 v
C. 2.03 V
D. -2.03 V
The voltage of an electrolytic cell with copper and magnesium electrodes is 2.71 v. Therefore, option B is correct.
What is an electrolytic cell ?The electrolytic cell is a type of cell that performs a redox reaction while using electrical energy. When electrical energy is applied, a redox reaction occurs in molten NaCl. It is therefore an electrolytic cell.
An electrolyte, two electrodes, and an electrolytic cell make up an electrolytic cell (a cathode and an anode). The electrolyte is typically a mixture of ions that have been dissolved in water or another solvent. Electrolytes can also be molten salts, like sodium chloride.
The standard reduction potential, E° of the metals are as below:
Mg²⁺ + 2e⁻ ⇌ Mg; E° = -2.372
Cu²⁺ + 2e⁻ ⇌ Cu; E° = +0.337
Therefore, magnesium has the the lower E°, it will serve as the anode in the electrolytic cell while copper will serve as the cathode.
At the anode; Mg ⇌ Mg²⁺ + 2e⁻, E° = -2.372
At the cathode; Cu²⁺ + 2e⁻ ⇌ Cu, E° = +0.337
EMF of the cell = E° cathode - E° anode
= 0.337 - (-2.372) = 2.71 V
Therefore, EMF of the cell is 2.71 V
Thus, option B is correct.
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a) A molecule of DNA contains 140 A bases, how many T bases will it contain? Plz help ❤️
Answer:
140 T or thymine base
Explanation:
Adenine pairs with Thymine in DNA thus the number of adenine will always equal number of thymine (unless some sort of mutation), therefore in this problem you have 140 A so you have 140 T as well. Remember: Adenine (A) and Thymine(T) is equal, & Cytosine (C) and Guanine (G) is equal
What type of bond will be formed for atoms that have a +1 or -1 charge?
Give the IUPAC name for the following compound: Multiple Choice (1R,3R)-1-ethyl-3-methylcyclohexane (1S,3S)-1-ethyl-3-methylcyclohexane (1R,3S)-1-ethyl-3-methylcyclohexane (1S,3R)-1-ethyl-3-methylcyclohexane
Answer:
(1R,3R)-1-ethyl-3-methylcyclohexane.
Explanation:
NOTE: The question is not complete since we do not have the diagram to the chemical structure in the question. Kindly check the attached picture for the diagram of the chemical structure.
So, in order to name Enantiomers or chemical structure through the use of the R,S system requires series of rules and regulations to follow for the proper naming.
There is an ethyl attached to the compound as the first substituents and methyl at the third which are the secondary prefix.
=> The longest chain is 6, thus the compound has hexane as the root compound.
=> It is (1R,3R) because when we draw from the highest substituents to the lower substituents, this is done in a clockwise direction.
A salt solution was found to contain 1.50 g of salt dissolved in 50 mL of water. On evaporation, the recovered salt weighed 1.47 g. What percent of salt was recovered?
A) 20.4%
B) 107%
C) 98%
D) 20.0%
Answer:
C = 98%
Explanation:
Hello,
To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.
Mass of salt recovered = 1.47g
Initial mass of salt = 1.50g
Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100
Percentage of salt recovered = (1.47 / 1.50) × 100
Percentage of salt recovered = 0.98 × 100
Percentage of salt recovered = 98%
The percentage of salt recovered is equal to 98%
To determine the concentration of a sample of calcium hydroxide, 1.45M HCl is added drop-wise using a burst. Write the balanced net ionic equation for the acid-base reaction.
Answer:
H^+(aq) + OH^-(aq) —> H2O(l)
Explanation:
We'll begin by writing the balanced equation for the reaction.
2HCl(aq) + Ca(OH)2(aq) —> CaCl2(aq) + 2H2O(l)
Ca(OH)2 is a strong base and will dissociates as follow:
Ca(OH)2(aq) —> Ca^2+(aq) + 2OH^-(aq)
HCl is a strong acid and will dissociates as follow:
HCl(aq) —> H^+(aq) + Cl^-(aq)
Thus, In solution a double displacement reaction occurs as shown below:
2H^+(aq) + 2Cl^-(aq) + Ca^2+(aq) + 2OH^-(aq) —> Ca^2+(aq) + 2Cl^-(aq) + 2H2O(l)
To get the net ionic equation, cancel out Ca^2+ and 2Cl^-
2H^+(aq) + 2OH^-(aq) —> 2H2O(l)
H^+(aq) + OH^-(aq) —> H2O(l)
A compound has an empirical formula of CHN. What is the molecular formula, if it’s molar mass is 135.13 g/mol? (C=12.01 amu, H=1.008 amu, N= 14.01)
Answer:
well the MF is 224.78 g/mol
Explanation:
just times them all by the molor mass and divide it by 3
MnCl₂ (aq) + (NH₄)₂CO₃ (aq) --> MnCO₃ (s) + 2 NH₄Cl (aq)
this equation is balanced.
Answer:
Nothing to do, is already balanced.
Explanation:
MnCl₂ (aq) + (NH₄)₂CO₃ (aq) --> MnCO₃ (s) + 2 NH₄Cl (aq)
Left Side
Mn =1
Cl = 2
N = 2
H = 8
C = 1
O = 3
Right Side
Mn =1
Cl = 2
N = 2
H = 8
C = 1
O = 3
g Identify the type of reaction [(combination (c), decomposition (d), combustion (co), single replacement (sr), double replacement (dr) , or neutralization (n)] and write a balanced equation. If there is no reaction, write NR. Aqueous ammonium phosphate reacts with aqueous calcium sulfide.
Answer: [tex]2(NH_4)_3PO_4(aq)+3CaS(aq)\rightarrow Ca_3(PO_4)_2(aq)+3(NH_4)_2S(s)[/tex] is a double displacement reaction (dr).
Explanation:
A double displacement reaction (dr) is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
[tex]2(NH_4)_3PO_4(aq)+3CaS(aq)\rightarrow Ca_3(PO_4)_2(aq)+3(NH_4)_2S(s)[/tex]
Combination reaction (C) is defined as the reaction where substances combine in their elemental state to form a single compound.
Single displacement reaction (sr) is defined as the reaction where more reactive element displaces a less reactive element from its chemical reaction.
Decomposition reaction (d) is defined as the reaction where a single substance breaks down into two or more simpler substances.
Combustion (Co) is a type of chemical reaction in which hydrocarbons burn in the presence of oxygen to form carbon dioxide and water along with heat.
In Chapter 4, we will learn that single bonds experience free rotation at room temperature, while double bonds do not. Consider the two C-N bonds in the structure. One of these bonds exhibits free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. Identify the C-N bond with restricted rotation, and justify your answer by drawing resonance structures.
Answer:
Explanation:
The main objective here is to draw a diagram of an heterocyclic compound containing two C-N bonds in the structure. One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. After that ; we will identify the C-N bond with restricted rotation, and also justify our answer by drawing resonance structures.
So; the first image below shows the structure of the heterocyclic compound containing two C-N bonds in the structure with One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. From the first diagram. the squared area indicates the C-N bond that exhibits restricted rotation.
The amide bonds in the C-N bonds offers the resonance characteristics and thus exhibits restricted rotation. The resonance is shown in the second image below
