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3 years ago

Consider the following balanced chemical equation representing the combustion of propane gas:

Chemistry

1 answer:

lora16 [44]3 years ago

6 0

Answer:

We are given the following equation:

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

we can see that the number of moles of H2O is 4/3 time the moles of CO2

So, the easiest way to find the number of moles of H2O is to find the number of moles of CO2 and multiply it by 4 / 3

Moles of CO2:

Molar mass of CO2 = 44 grams / mol

Moles of CO2 = Given mass / Molar mass

Moles of CO2 = 16.3 / 44 moles

Moles of H2O:

Moles of H2O = Moles of CO2 * 4 / 3

Moles of H2O = 16.3 * 4 / 3 * 44

Moles of H2O = 16.3 / 3 * 11

Moles of H2O = 16.3 / 33

Moles of H2O = 0.5 moles (approx)

Therefore, along with 16.3 grams of CO2, 0.5 moles of H2O will also be formed

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0.7 moles of uranium to grams of uranium

alexandr1967 [171]

Answer: 166.6g

Explanation: 0.7mol x 238g/1mol

8 0

3 years ago

How many bonded pairs of electrons are there in the molecule H2CO?

Drupady [299]

There are eight bonded electrons in this molecule! :)

8 0

4 years ago

The hydrogen ion concentration, [H+], in a certain cleaning compound is left bracket Upper H Superscript plus Baseline right

labwork [276]

Answer: The pH of the cleaning compound is 10.44

Explanation:

pH is defined as the negative logarithm of hydrogen or hydronium ion concentration that are present in a solution.

The equation representing pH of the solution follows:

$pH=-\log[H^+]$

We are given:

$[H^+]=3.6\times 10^{-11}$

Putting values in above equation, we get:

$pH=-\log(3.6\times 10^{-11})$

$pH=10.44$

Hence, the pH of the cleaning compound is 10.44

8 0

3 years ago

which solution has a higher percent ionization of the acid , a .10 M solution of HC2H3O2 (aw) or a .010 M solution of HC2H3O2(aq

ruslelena [56]

Answer:

0.0010 M of HC2H302(aq)

3 0

3 years ago

A mixture of sodium bicarbonate and ammonium bicarbonate is 75.9% bicarbonate by mass. what is the mass percent of sodium bicarb

Lorico [155]

Answer: 24.1%, under below assumptions.

Justification:

The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:

- sodium bicarbonate, and
- ammonium bicarbonate

.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.

Given that later, the question is what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.

With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.

In formulas:

100% = %m/m sodium bicarbonate + %m/m ammonium bicarbonate = 100%

=> % m/m sodium bicarbonate = 100% - % m/s ammonium bicarbonate

=> % sodium bicarbonate = 100% - 75.9% = 24.1%

Answer: 24.1%

8 0

4 years ago