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kondaur [170]
3 years ago
5

Need helpppppp hurry pls, i just really need help

Chemistry
1 answer:
mojhsa [17]3 years ago
7 0

Answer:

6.8 mole of O₂

Explanation:

Given expression:

       2H₂   +   O₂   →   2H₂O

Number of moles of H₂ = 13.6moles

Unknown:

Number of moles of O₂  = ?

Solution:

In the given problem, we are to find the number of moles of oxygen gas that will use up 13.6mole of hydrogen gas;

 From the reaction equation;

             2 mole of H₂ will completely react with 1 mole of O₂

         13.6 moles of H₂ will completely be used up by \frac{13.6}{2} mole of  O₂

                                                                                        = 6.8 mole of O₂

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Answer:

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Explanation:

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A sample of methane collected when the temp was 30 C mmHg measures 398 mL. What would be the volume of the sample at -5 C and 61
Levart [38]
<h2>Question </h2>

A sample of methane collected when the temp was 30 C and 760mmHg measures 398 mL. What would be the volume of the sample at -5 C and 616 mmHg pressure

<h2>Answer:</h2>

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<h2>Explanation:</h2>

Using the combined gas law:

\frac{PV}{T} = k

Where;

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V = Volume

T = Temperature

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It can be deduced that:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = k           ---------------------(i)

Where:

P₁ and P₂ are the initial and final pressures of the given gas

V₁ and V₂ are the initial and final volumes of the given gas

T₁ and T₂ are the initial and final temperatures of the gas.

<em>From the question:</em>

the gas is methane

P₁  = 760mmHg

P₂ = 616mmHg

V₁ = 398mL

V₂ = ?

T₁ = 30°C = (30 +273)K = 303K

T₂ = -5°C = (-5 +273)K = 268K

Substitute these values into equation (i) as follows;

\frac{760*398}{303} = \frac{616*V_2}{268}

Solve for V₂

V₂ = \frac{760*398*268}{616*303}

V₂ = 434.32mL

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B. i think not 100% sure 
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How molecules react to an increase in temperature.
Slav-nsk [51]

Answer:

An increase in temperature typically increases the rate of reaction. An increase in temperature will raise the average kinetic energy of the reactant molecules. Therefore, a greater proportion of molecules will have the minimum energy necessary for an effective collision (Figure. 17.5 “Temperature and Reaction Rate”).

Explanation:

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3 years ago
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