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4 years ago

Why does an increase in reaction temperature generally increase the reaction rate?

Chemistry

2 answers:

GrogVix [38]4 years ago

6 0

Answer:

cuz of the disproportionately big increase in the number of high energy collisions.

zalisa [80]4 years ago

5 0

Answer:

This is because an increase in temperature increases the movement of molecules/particles of reactants which encourages collision and subsequently, reaction.

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Fill in the blanks with the words given below- [Atoms, homogeneous, metals, true, saturated, homogeneous, colloidal, compounds,

Virty [35]

Answer:

1. metals

2. atom

3. homogeneous

4. compounds

5. lustrous

6. saturated

7. colloidal

8. homogeneous

Explanation:

5 0

3 years ago

A(n) ________is your legal document issued by a government that gives an inventor exclusive rights to make use or sell intervent

Crazy boy [7]

Patent - Not sure if this is much of a Chem question dou.

7 0

4 years ago

Read 2 more answers

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g$

Theoretical yield of $CO_2$ = 0.572 g

Experimental yield of $CO_2$ = 0.391 g

Now we have to calculate the percent yield of $CO_2$

$\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100$

$\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%$

Therefore, the percent yield of $CO_2$ is, 68.4 %

6 0

3 years ago

Which statement best describes the revolution of a planet? A. the Earth's orbit around the sun B. the time it takes the planet t

ollegr [7]

I would d hope this helps

3 0

4 years ago

Read 2 more answers

A sample of gas (1.3 mol) occupies __________ l at 22 Â°c and 2.5 atm.

Furkat [3]

To answer this question, you just need to put all the variable into PV = nRT formula. In this case, the pressure(P) is 2.5 atm, the number of molecule(n) is 1.3 mol, the temperature(T) is 22 celcius= 295.15 kelvin.
Then the volume should be:

PV = nRT
2.5 * V = 1.3 * 0.08206* 295.15
V= 12.59L

4 0

3 years ago