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2 years ago

Does this chemical equation satisfy the Law of Conservation of Mass? CS 2 +O 3 -->CO 2 +SO 2

Chemistry

1 answer:

makkiz [27]2 years ago

4 0

Answer:

No it doesn't satisfy the Law of Conservation of Mass.

Explanation:

A balanced chemical equation obeys the law of conservation of mass.

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Which state of matter is Na OH(s)

Shalnov [3]

Explanation:

SOLID

Sodium hydroxide exists in the solid phase at room temperature. You would find it in the lab as hemispherical white solid pellets. The phase of a substance depends on temperature and pressure. As you heat a solid, it will melt and change to the liquid phase.

4 0

3 years ago

How much heat is required to increase the temperature of 198.5 grams of water from 25.0

velikii [3]

5.27*10^4 (52,700.)

4 0

3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

3 years ago

Write a balanced chemical equation for the standard formation reaction of gaseous hydrogen fluoride hf

pychu [463]

The standard state formation reaction is a chemical reaction in which one moles of substance in its standard state is formed from its constituent element in their standard state.All the substance must be in their most stable state at 100kpa and 25 degrees celsius.
therefore for HF is
1/2H2 +1/2F2 =HF

6 0

3 years ago

Read 2 more answers

What pressure will be produced when 2.0 moles of N2 gas is heated to 68oC in a container that holds 1.25 of gas?

Stella [2.4K]

The pressure of the nitrogen gas produced is determined as 44.77 atm.

<h3>What is the pressure of the Nitrogen gas?</h3>

The pressure of the nitrogen gas is determined from ideal gas equation, as shown below;

PV = nRT

P = nRT/V

where;

n is number of moles = 2 moles
R is ideal gas constant = 0.08205 L.atm/mol.K
T is temperature = 68⁰C = 68 + 273 = 341 K
V is volume = 1.25 L

P = (2 x 0.08205 x 341)/(1.25)

P = 44.77 atm.

Learn more about pressure here: brainly.com/question/25736513

#SPJ1

3 0

2 years ago