How to balance s+hno3=h2so4+no2+h2o

klemol [59]

I can’t see the choices can you take another picture of this assignment please so i can help you

6 0

2 years ago

A 0.100 M solution of K2SO4 would contain the same total ion concentration as which of the following solutions? 0.0800 M Na2CO3

mr Goodwill [35]

0.0750 M Na3PO4 as this solution would contain 3 Na+ and 1 PO4- ions per mole of Na3PO4 for an effective total ion concentration of 4 x .0750 or .300 M. The K2SO4 has three total ions or a concentration of .300 M as well. Hope it helps.

6 0

3 years ago

Sunlight strikes Earth’s surface at different angles This angle is called the angle of

stira [4]

Answer:

Earth takes in thermal energy from the Sun in a process called. ... Sunlight strikes Earth's surface at different angles. This angle is called the angle of. ⇒ insolation.

Explanation:

Report the guy that spammed, he spams for points and doesn't even try to help.

4 0

2 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

3 years ago

A drinking water plant adds 500 grams of fluoride to a water tank containing 500,000 liters of drinking water. what is the conce

nlexa [21]

Answer is: concentration of fluoride in the water in parts-per-million is 1 ppm.
Parts-per-million (10⁻⁶) is present at one-millionth of a gram per gram of sample solution, for example mg/kg.
m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.

5 0

2 years ago