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3 years ago

Bleach is a 0.750 M solution of sodium hypochlorite. What volume, in litres, would contain 120.3 g of sodium hypochlorite?

Chemistry

1 answer:

Lady bird [3.3K]3 years ago

4 0

Answer:

2.15 L

Explanation:

M(NaOCl) = 23.0 + 16.0 + 35.5 = 74.5 g/mol

120.3 g *1 mol/74.5 g = 1.615 mol NaOCl

0.750 M = 0.750 mol/L

1.615 mol * 1L/0.750 mol = 2.15 L

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At equilibrium the partial pressures of N2O4 and NO2 are 0.35 atm and 4.3 atm. What is the Kp

ale4655 [162]

Answer:

Kp = 52.83

Explanation:

6 0

3 years ago

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago

5. What is the mass of 9.80 x 1023 formula units of zinc chlorate, Zn(CO3)2?

Zielflug [23.3K]

Answer:

$\boxed{\text{378 g}}$

Explanation:

We must convert formula units of Zn(ClO₃)₂ to moles and then to grams of Zn(ClO₃)₂.

Step 1. Convert formula units to moles

$\text{Moles of Zn(ClO$_{3}$)$_{2}$}\\\\= 9.80 \times10^{23}\text{ formula units Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}{6.022 \times\ 10^{23} \text{ formula units Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$}$

Step 2. Convert moles to grams

$\text{Mass of Zn(ClO$_{3}$)$_{2}$}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{232.29 g Zn(ClO$_{3}$)$_{2}$}}{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{378 g Zn(ClO$_{3}$)$_{2}$}\\\\\text{The mass of Zn(ClO$_{3}$)$_{2}$ is } \boxed{\textbf{378 g}}$

5 0

3 years ago

Chemistry: Molarity and Concentrations

kotegsom [21]

I dont know if this correct, but i hope it help

5 0

2 years ago

What accounts for the large increase in volume when sugar burns?

Dimas [21]

I don’t know type it up yourself

8 0

3 years ago