Answer:
u
uric acid is a useful diagnostic tool as screening for most of purine metabolic disorders. The importance of uric acid measurement in plasma and urine with respect of metabolic disorders is highlighted. Not only gout and renal stones are indications to send blood to the laboratory for uric acid examination
Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is given by the equation 4C2H5O2N(s) + 9O2(g) → 8CO2(g) + 10H2O(l) + 2N2(g) ΔH°rxn = –3857 kJ/mol Given that ΔH°f[CO2(g)] = –393.5 kJ/mol and ΔH°f[H2O(l)] = –285.8 kJ/mol, calculate the enthalpy of formation of glycine.
Answer:
ΔH°f C₂H₅O₂N(s) = -537.2kJ
Explanation:
Based on the reaction:
4 C₂H₅O₂N(s) + 9O₂(g) → 8CO₂(g) + 10H₂O(l) + 2N₂(g)
ΔHrxn = ΔH°f products - ΔH°f reactants.
As:
ΔH°fO₂(g) = 0
ΔH°fCO₂(g) = -393.5kJ/mol
ΔH°fH₂O(l) = -285.8kJ/mol
ΔH°fN₂(g) = 0
The ΔHrxn is:
ΔHrxn = (8×-393.5kJ/mol + 10×-285.8kJ/mol) - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-6006kJ/mol - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol
-4×ΔH°fC₂H₅O₂N(s) = 2149kJ/mol
ΔH°fC₂H₅O₂N(s) = 2149kJ/mol / -4
ΔH°f C₂H₅O₂N(s) = -537.2kJA gas has a volume of 6.6 L at a temperature of 40 C. What is the volume of
the gas if the temperature changes to 15 C?
Answer:
6.07 L
Explanation:
It appears that the reading has been made at constant pressure .
At constant pressure , the gas law formula is
V/T = constant V is volume and T is temperature of the gas.
V₁ / T₁ = V₂ / T₂
V₁ = 6.6 L ,
T₁ = 40°C
= 273 + 40
= 313 K
T₂ = 15+ 273
= 288K
V₂ = ?
Putting the values in the formula above
6.6 / 313 = V₂ / 288
V₂ = 6.07 L.
a binary ionic compound is made of two components name one of them
Answer:
CATION
Explanation:
It's one is the action and the mother is a cation.
What is Hess‘s law please help
The correct answer is D. Hess's law states than the enthalpy of a reaction does not depend on the reaction path
Explanation:
In a chemical reaction, the enthalpy refers to the internal energy in a system and how this increases or decreases during the reaction. According to Hess's law proposed by German Hess in 1940, the enthalpy does not depend on the reaction path or the number of steps in a reaction. This means one reaction of only one step will have the same enthalpy that if the reaction occurs in several steps because the energy that requires all the process is the same. Thus, the "Hess's law states than the enthalpy of a reaction doe s not depend on the reaction path".
Consider this reaction:
2Cl2O5 —> 2Cl2 + 5O2
At a certain temperature it obeys this rate law.
rate = (2.7.M^-1•s^-1) [Cl2O5]^2
Suppose a vessel contains Cl2O5 at a concentration of 0.600M. calculate how long it takes for the concentration of Cl2O5 to decrease by 94%. you may assume no other reaction is important. round your answer to two digits
Answer:
[tex]t=9.7s[/tex]
Explanation:
Hello,
In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:
[tex]\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}[/tex]
Thus, the final concentration for a 94% decrease is:
[tex][Cl_2O_5]=0.600M-0.600M*0.94=0.036M[/tex]
Therefore, we compute the time for such decrease:
[tex]kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}[/tex]
[tex]t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s[/tex]
Regards.
Photochromic lenses contain Group of answer choices both AgCl and CuCl embedded in the glass. only AgCl embedded in the glass. neither AgCl nor CuCl embedded in the glass. only CuCl embedded in the glass.
Answer:
both AgCl and CuCl embedded in the glass
Explanation:
Photochromic lenses contain both AgCl and CuCl embedded in the glass.
They are light-sensitive lenses that adapt to environmental changes. They appear clear when in an apartment or a building and automatically darken when outside as a result of exposure to sunlight. The darkening is activated by the UV component of the sunlight.
Photochromic lenses are otherwise known as light-adaptive or intelligent lenses and they are formed by coating lenses with silver chloride compounds whose concentration ranges from 0.01 to 0.001 %. Copper (I) chloride is also included in addition to the silver halide.
In summary, photochromic lenses contain both AgCl and CuCl.
Discuss any give ways by which
the falling moral standards of Ghanaian
youth can be minimised.
Answer:
The falling standards of Ghanaian youths can be minimized by proper upbringing of the children by their parents. The youths should be taught about what is wrong or right and there should be a corresponding reward for those who do good and exceptional in order to encourage others in towing that line and punishment should also be meted out to those who break the law. Mediocrity shouldn’t be celebrated and the elders should lead by example.
These will make the falling standards of Ghanaian youth get reduced.
Hcl and 1-isopropylcyclohexane formation
Answer:
Spahgetti
Explanation:
what is the chemical symbol and name of the third element in the periodic table
Answer: Aluminum symbol Al or aluminum American English
Explanation:
Answer:
Hii
Li( Lithium)
Explanation:
Lithium has the atomic number of three and is the third element in periodic table.
Round off the following measurement to three significant digits: 29.950g
Answer:
30.0 g.
Explanation:
Hello,
In this case, for us to round the given number off to three significant figures, we firstly realize it has initially five significant figures. Thus, cutting at the third digit, which is the second nine, we will have 29.9 g, nonetheless, as a five is after such nine, we should round the nine to ten, so the result is 30.0 g.
Best regards.
How many moles of H2 are needed to produce 34.8 moles of NH3?
2 i hope this helps
:)✨✨✨✨✨✨
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?
Aspirin is usually packaged with
A. acetic anhydride
B. salicylic acid
C. buffering agents
Answer:
Aspirin is usually packaged with C. buffering agents.
Explanation:
In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for electrons. II. the oxidized form has a lower affinity for electrons. III. the reduced form has a higher affinity for electrons. IV. the greater the tendency for the oxidized form to accept electrons.
Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
A pure sample of the R enantiomer of a compound has a specific rotation, [ α], of +20 °. A solution containing 0.2 g/mL of a mixture of enantiomers rotates plane polarized light by −2 ° in a 1 dm polarimeter. What is the enantiomeric excess (%ee) of the mixture?
Answer:
Explanation:
The specific rotation of the sample is -2 degrees/0.2 g/mL of mixture
This equals -10 degrees/g/mL of sample.
let the proportion of the R (+) enantiomer be x. The proportion of the S (-) enantiomer in the mixture will be given by (1-x).
specific rotation of the mixture = proportion of R enantiomer* its specific rotation + proportion of S enantiome * its specific rotation
i.e.
-10 = x *(+20) + (1-x)*(-20)
-10 = 20x-20 + 20x
-10+20 = 40x
+10 = 40 x
x=10/40 = 25%
Since the proportion of the other enantiomer is 1-x, it is 0.75 or 75%
So the mixture contains 25% R, 75% S, giving you an excess of 50%.
Answer:
10%
Explanation:
Enantiomeric excess is a way of describing how optically pure a mixture is by calculating the purity of the major enantiomer. It can range from 0%-100%. Enantiomeric excess ( ee ) can also be defined as the absolute difference between the mole fractions of two enantiomers.
Enantiomeric excess is also called optical purity. This is because chiral molecules cause the rotation of plane-polarized light and are said to be optically active. An enantiomerically pure sample has an enantiomeric excess of 100 percent
Enantiomeric excess = observed specific rotation/specific rotation of the pure enantiomer x 100
From the data given in the question;
observed specific rotation= -2°
specific rotation of the pure enantiomer = +20°
Therefore;
ee= 2/20 ×100
ee= 10%
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?
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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The major source of aluminum in the world this bauxite (mostly aluminum oxide). It’s thermal decomposition can be represented by:
Al2 O3 (s) —> 2 Al (s) + 3/2 O2 (g)
ΔH rxn = 1676
If aluminum is produced this way, how many grams of aluminum can conform when 1.000×10^3 kJ of heat is transferred?
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
What is a thermochemical equation?A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change.
Step 1: Write the thermochemical equation.Al₂O₃(s) ⇒ 2 Al(s) + 3/2 O₂(g) ΔH rxn = 1676 kJ
Step 2: Calculate the moles of Al formed when 1.000 × 10³ kJ of heat is transferred.According to the thermochemical equation, 2 moles of Al are formed when 1676 kJ of heat is transferred.
1.000 × 10³ kJ × (2 mol Al/1676 kJ) = 1.193 mol Al
Step 3: Calculate the mass corresponding to 1.193 moles of AlThe molar mass of Al is 26.98 g/mol.
1.193 mol × 26.98 g/mol = 32.19 g
In the thermal decomposition of aluminum oxide, the transference of 1.000 × 10³ kJ of heat can produce 32.19 g of Al.
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Describe why some acids are strong while other acids are weak
Answer:
I hope this help you. Mark me as brainliest and rate pleaseExplanation:
the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.
It is important that you don't confuse the words strong and weak with the terms concentrated and dilute.
As you will see below, the strength of an acid is related to the proportion of it which has reacted with water to produce ions. The concentration tells you about how much of the original acid is dissolved in the solution.
It is perfectly possible to have a concentrated solution of a weak acid, or a dilute solution of a strong acid.
Chemical formula for copper gluconate I have 1.4g of Copper gluconate. There is .2g of copper within the copper gluconate. Determine the chemical formula for Copper gluconate with the given information: Copper Gluconate: Cu(C6H11O?)? Cu = 63.55 g/mol H = 12.01 g/mol O = 1.008 g/mol Cu = 63.55 g/mol
Answer:
The simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂
Explanation:
Given mass of sample = 1.4 g
mass of copper in the sample = 0.2 g
mass of the gluconate =1.4 - 0.2 = 1.2 g
The mole ratio is determined first using the formula;
mole ratio = reacting mass / atomic mass
atomic mass of copper = 63.55
mass of gluconate, C₆H₁₁O₇ = 12*6 + 1*11 + 16*7 = 195 g/mol
mole ratio ( copper : gluconate) = 0.2/63.55 : 1.4/195
mole ratio ( copper : gluconate) = 0.003 : 0.007
convert to whole number ratios by dividing with the smallest ratio
mole ratio ( copper : gluconate) = 0.003/0.003 : 0.007/0.003
mole ratio ( copper : gluconate) = 1 : 2
Therefore, the simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂
List three ways the rate of solvation of sodium chloride in water may be
increased
Answer:
1) Increasing temperature
2) Stirring
3) Increasing surface area of salt by grinding it
What is the molar mass of CH2O2 ? ( C= 12.01 g/mol, H=1.008 g/mol, O=16.00)
Answer:
Molar Mass of CH2O2 is 46.026
Explanation:
What is the molar mass of CH2O2 ? ( C= 12.01 g/mol, H=1.008 g/mol, O=16.00)
C = 12.01g/mol
H = 1.008g/mol
O = 16g/mol
CH2O2 = 12.01+1.008x2+16x2 = 46.026g/mole
Ba(OH)2:_______.
A. 1 barium atom, 1 oxygen atom and 1 hydrogen atom.
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
C. 1 barium atom, 2 oxygen atoms and 2 hydrogen atoms.
D. 1 barium atom, 2 oxygen atoms and 1 hydrogen atom.
Answer: D
Explanation: Expand this (OH)2 you will get 2O, 2H
Hence 1Ba, 2O, 2H
Answer:
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
Which of the following is an example of a mechanical wave?
O A. A light ray
B. A seismic wave
C. A radio wave
D. An X-ray
Answer:
A seismic wave
Explanation:
It requires a medium for its propagation.
From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O
How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g
Include the correct number of significant figures in your final answer
Answer: 125 g
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]
The balanced reaction is:
[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]
According to stoichiometry :
1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex] of [tex]O_2[/tex]
Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]
Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]
A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 44./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 24.01g water 13.10g Use this information to find the molecular formula of X.
Answer:
C3H6.
Explanation:
Data obtained from the question:
Mass of the compound = 8g
Mass of CO2 = 24.01g
Mass of H2O = 13.10g
Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 44g/mol
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01
=> 12/44 x 24.01 = 6.5g
Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1
=> 2x1/18 x 13.1 = 1.5g
Mass of O in the compound = Mass of compound – (mass of C + Mass of H)
=> 8 – (6.5 + 1.5) = 0
Next, we shall determine the empirical formula of the compound. This is illustrated below:
C = 6.5g
H = 1.
Divide by their molar mass
C = 6.5/12 = 0.54
H = 1.4/1 = 1.
Divide by the smallest
C = 0.54/0.54 = 1
H = 1/0.54 = 2
Therefore, the empirical formula is CH2
Finally, we shall determine the molecular formula as follow:
The molecular formula of a compound is a multiple of the empirical formula.
Molecular formula = [CH2]n
[CH2]n = 44
[12 + (2x1)]n = 44
14n = 44
Divide both side by 14
n = 44/14
n = 3
Molecular formula = [CH2]n = [CH2]3 = C3H6
Therefore, the molecular formula of the compound is C3H6
The calculated yield for the production of carbon dioxide was 73.4g. When the
experiment was performed in the lab, a yield of 72.3g was produced. What is the
percent yield of carbon dioxide?
Answer:10 grams of CO2
Explanation:
Yeild= exp. yeild÷ thoretical yeild × 100
Yeild= 73.3÷73.4 × 100
Yeild= 0.1 ×100
Yeild= 10
What is the equilibrium constant for the following reaction:HCO2H(aq) + CN–(aq) HCO2–(aq) + HCN(aq)Does the reaction favor the formation of reactants or products? The acid dissociation constant, Ka, for HCO2H is 1.8 x 10–4and the acid dissociation constant for HCN is 4.0 x 10–10.(A) K = 1.00. The reaction favors neither the formation of reactants nor products.(B) K = 2.2 x 10–6. The reaction favors the formation of products.(C) K = 2.2 x 10–6. The reaction favors the formation of reactants.(D) K = 4.5 x 105. The reaction favors the formation of products.(E) K = 4.5 x 105. The reaction favors the formation of reactants.
Answer:
(D) K = 4.5 x 10⁵. The reaction favors the formation of products
Explanation:
HCOOH + CN⁻ ⇆ HCOO⁻ + HCN
K = [HCOO⁻] [ HCN ] / [ HCOOH] [ CN⁻]
HCOOH ⇄ H ⁺ + COO⁻
K₁ = [ H⁺] [ COO⁻ ] / [HCOOH ]
HCN ⇆ H⁺ + CN⁻
K₂ = [ H⁺] [ CN⁻] / [ HCN ]
K₁ / K₂
= [ H⁺] [ COO⁻ ] / [HCOOH ] X [ HCN ] / [ H⁺] [ CN⁻]
= [ COO⁻ ][ HCN ] / [HCOOH ] [ CN⁻]
= K
K = K₁ / K₂
= 1.8 x 10⁻⁴ / 4 x 10⁻¹⁰
= 4.5 x 10⁵
So equilibrium constant of the reaction
HCOOH + CN⁻ ⇆ HCOO⁻ + HCN
is very high . Hence reaction favours the formation of product.
option (D) is correct.
An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted to a total solution volume of 2.00 L. Calculate the molarity, molality, and mass percent of the solution. (Assume the density of 1.05 g/mL for the solution.)
Answer:
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of the solution es 3.45%
Explanation:
Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:
[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Being:
K: 39 g/moleN: 14 g/moleO: 16 g/moleThe molar mass of KNO₃ is:
KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole
You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?
[tex]moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}[/tex]
moles of KNO₃= 0.718
So you have:
moles of KNO₃= 0.718volume= 2 LApplying this quantity in the definition of molarity:
[tex]molarity=\frac{0.718 moles}{2 L}[/tex]
Molarity= 0.359[tex]\frac{moles}{L}[/tex]
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.
Then the molality is calculated by:
[tex]Molality=\frac{moles of solute}{mass of solvent in kilograms}[/tex]
Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?
[tex]mass=\frac{2000 mL*1.05 grams}{1 mL}[/tex]
mass= 2100 grams
Since mass solution = mass water + mass KNO₃
then mass water = mass solution - mass KNO₃
Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams
mass water= 2,027.5 grams
Then, being:
moles of KNO₃= 0.718mass of solvent in kilograms= 2.0275 kg (being 2,027.5 grams= 2.0275 kilograms because 1,000 grams= 1 kilogram)Replacing in the definition of molality:
[tex]molality=\frac{0.718 moles}{2.0275 kg}[/tex]
molality= 0.354 [tex]\frac{moles}{kg}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.
[tex]mass percent=\frac{mass of solute}{mass of solution} *100[/tex]
So, in this case:
[tex]mass percent=\frac{72.5 grams}{2100 grams} *100[/tex]
mass percent= 3.45 % KNO₃ by mass
The mass percent of the solution es 3.45%
The molarity of the solution is 0.36 mol/L. The molality of the solution is 0.34 m. The mass percent of the solution is 3.33%.
Number of moles of KNO3 = mass/molar mass = 72.5 g/101 g/mol = 0.72 moles
Molarity = Number of moles / volume = 0.72 moles/ 2.00 L = 0.36 mol/L
The molality = Number of moles of solute/Mass of solution in kilograms
mass of solution = 1.05 g/mL × 2000 mL = 21000 g or 2.1Kg
Molality of solution = 0.72 moles/2.1 Kg = 0.34 m
Mass percent of solution = mass of solute/mass of solution × 100/1
Mass percent of solution = 72.5 g/ (72.5 g + 2100 g) × 100/1
= 3.33%
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The equilibrium constant for the reaction NO2(g)+NO3(g)→N2O5(g) is 2.1x10-20 , therefore: a. At equilibrium, the concentration of products and reactants is about the same. b. At equilibrium, the concentration of products is greater than the reactants. c. At equilibrium, the concentration of reactants is greater than the products
Answer: c. At equilibrium, the concentration of reactants is greater than the products
Explanation:
Equilibrium constant for a reaction is the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.
For the reaction:
[tex]NO_2(g)+NO_3(g)\rightleftharpoons N_2O_5(g)[/tex]
Equilibrium constant is given as:
[tex]K_{eq}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}[/tex]
[tex]2.1\times 10^{-20}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}[/tex]
When
a) K > 1, the concentration of products is greater than the concentration of reactants
b) K < 1, the concentration of reactants is greater than the concentration of products
c) K= 1, the reaction is at equilibrium, the concentration of reactants is equal to the concentration of products
Thus as [tex]K_{eq}[/tex] is [tex]2.1\times 10^{-20}[/tex] which is less than 1,
the concentration of reactants is greater than the concentration of products
Benzene can be converted to 1,3,5-tribromobenzene in five reaction steps and four intermediate compounds. Select the appropriate reagent from the followings.
Br2, R2O2
CH3Cl, AlCl3
CH3COCl, AlCl3
NaNO2, HCl
HNO3, H2SO4
H3PO2
H3PO4
KMnO4
Answer:
The appropriate reagent is: H3PO2.
Explanation:
H3PO2 is in charge of eliminating the amino group by diazotization, remember that the amino group had previously achieved bromination at positions m; that is to say that it achieved in the beginning that the three bromine atoms of 1,2,4 tribromobenzene were introduced in the meta positions among themselves, which finally corresponds as part of the last reaction to the 1,3,5-tribromobenzene position.
