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snow_lady [41]
3 years ago
7

67. Name the elements contained in the following

Chemistry
1 answer:
Orlov [11]3 years ago
4 0
A. sodium and chlorine b. carbon and hydrogen c. nitrogen and hydrogen d. bromine
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Nitric acid (HNO3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.
Alborosie

<u>Given:</u>

Concentration of HNO3 = 7.50 M

% dissociation of HNO3 = 33%

<u>To determine:</u>

The Ka of HNO3

<u>Explanation:</u>

Based on the given data

[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M

The dissociation equilibrium is-

            HNO3   ↔    H+      +      NO3-

I            7.50               0                 0

C          -2.48          +2.48              +2.48

E            5.02            2.48              2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

6 0
3 years ago
1. Interpret the following equation using moles, molecules, and volumes (assume STP). Compare the mass of the reactants to the m
blsea [12.9K]
1. In this reaction, 2 moles of nitrogen gas reacts with 3 moles of oxygen gas to give 2 moles of N2O3 gas. 2 nitrogen molecules react with 3 oxygen molecules to give 2 N2O3 molecules.  Under STP, one mole of an ideal gas occupies a volume of 22.4 liters. So in this reaction, 44.8 liters of nitrogen gas reacts with 67.2 liters of oxygen gas to give 44.8 liters of N2O3 gas.  The total mass of the reactants (N2 and O2) is the same as the total mass of the product (N2O3). This is called mass balance of a chemical reaction.


2. According to the chemical reaction, 3 moles of chlorine gas produces 2 moles of iron(III) chloride.  So, to produce 1 moles of iron(III) chloride, 3/2 (1.5) moles of chlorine gas is required.  Therefore, to produce 14 moles of iron(III) chloride, 14 x 1.5 = 21 moles of chlorine is needed.
3 0
4 years ago
Read 2 more answers
Please help. im freaking out rn. i have like 40 missing assignments please
Katarina [22]

Answer:

I'm pretty sure its the one that says very little at the beginning but if I get it wrong I'm sorry

5 0
3 years ago
By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask
tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L

0.823 Molar is the molarity of the solution.

6 0
3 years ago
calculate the number of moles of sulfuric acid that is contained in 250 mL of 8.500 M sulfuric acid solution
san4es73 [151]

Answer : The moles of H_2SO_4 are, 2.125 mole.

Explanation : Given,

Molarity of H_2SO_4 = 8.500 M

Volume of solution = 250 mL  = 0.250 L    (1 L = 1000 mL)

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of }H_2SO_4}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

8.500M=\frac{\text{Moles of }H_2SO_4}{0.250L}

\text{Moles of }H_2SO_4=2.125mol

Therefore, the moles of H_2SO_4 are, 2.125 mole.

5 0
3 years ago
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