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Chemistry

1 answer:

Oliga [24]2 years ago

4 0

Answer:

Explanation:

Increasing the the temperature would favour the endothermic reaction which is the forward direction however increasing the pressure would make the reaction try to counteract this change by favouring the reaction that would create more products so the equilibrium will shift left instead of right.

Hope this helps.

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Climate is the __________________ average weather conditions that occur in a particular region.

masya89 [10]

Explanation:

Climate is the period of time average weather condition which occurs at a place

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2 years ago

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28.7 l of propane, c3h8, are consumed in a combustion reaction. how many grams of water are produced?

lora16 [44]

The chemical reaction would be:

C3H8 + 5O2 = 3CO2 + 4H2O

For this case, we assume that gas is ideal thus in every 1 mol the volume would be 22.41 L. We calculate as follows:

28.7 L C3H8 ( 1 mol / 22.41 L ) ( 4 mol H2O / 1 mol C3H8 ) ( 18.02 g / mol ) = 92.31 g H2O produced

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7 0

3 years ago

Electricity is the _of charged<br> particles.<br> A.movement<br> B. collection<br> C.build up

Vitek1552 [10]

Answer:

Explanation:

4 0

3 years ago

How many chlorine atoms are there in 12.2 g of ccl4? How many chlorine atoms are there in 12.2 of ? 4.77×1022 atoms?

mash [69]

Answer:- $1.91*10^2^3Cl$ atoms.

Solution:- We have been given the grams of carbon tetrachloride and asked to calculate the number of atoms of chlorine. It is a three step conversion problem. In the first we convert the grams of carbon tetrachloride to moles of it. In second step we convert moles of carbon tetrachloride to moles of chlorine and in the third step we convert the moles of chlorine to atoms of chlorine.

For grams to mole conversion we need the molar mass of the compound. Molar mass of carbon tetrachloride is 153.82 grams per mol. If we look at the formula of carbon tetrachloride then four chlorine are present in it. It means 1 mol of carbon tetrachloride has four moles of chlorine. The calculations are as follows:

$12.2gCCl_4(\frac{1molCCl_4}{153.82gCCl_4})(\frac{4molCl}{1molCCl_4})(\frac{6.02*10^2^3Clatoms}{1molCl})$