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brilliants [131]
3 years ago
5

Time (sec) [CH4] (mol/L)

Chemistry
1 answer:
Oxana [17]3 years ago
8 0

Based on the table and the equation of reaction; the

  • rate of consumption of oxygen between time 20 and 40s is 0.005 M/s
  • rate of consumption of oxygen between time 40 and 60s is 0.0025 M/s

<h3>What is rate of a reaction?</h3>

The rate of a reaction is the rate at which reactants molecules are consumed or products are formed.

  • Rate of reaction = change in concentration of reactants or products/ time taken

Based on the equation of reaction, the rate of consumption of oxygen is equal to the rate of consumption of methane.

Thus. rate of consumption of oxygen from time 20 seconds to time 60 seconds is calculated as follows:

Rate between 20 and 40 = 1.0 - 0.9/20

  • Rate of consumption of oxygen = 0.005 M/s

Rate between 40 and 60 = 0.9 - 0.85/20

  • Rate of consumption of oxygen = 0.0025 M/s

Therefore, the rate of consumption of oxygen decreases with time.

Learn more about rate of reaction at: brainly.com/question/25724512

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Answer:

Sodium carbonate

Explanation:

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How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group
OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

  • P= 1.00 atm
  • V= 150 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 \frac{atm*L}{mol*K} *273 K

Solving:

n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

mass=\frac{6.69 moles*70.9 grams}{1 mole}

mass= 474.321 grams ≅ 474 grams

<u><em>474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C</em></u>

4 0
3 years ago
9.69 × 10^7 in decimal form
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The answer is 96,900,000

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The enthalpy of neutralization for the reaction of a strong acid with a strong base is −56 kJ/mol of water produced. How much en
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Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and
blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g;     Hydrogen H = 2.24 g     Oxygen O = 71.1 g

To determine the empirical formula;

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For Hydrogen:

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Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon

For Hydrogen:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen

For Oxygen:

\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen

Thus, the empirical formula is HCO₂

4 0
3 years ago
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