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3 years ago

What mass of oxygen is needed for the complete combustion of 4.60×10−3g of methane?

1 answer:

7 0

First step in answering the question is to establish a balanced chemical reaction equation. More specifically, a combustion chemical equation.

CH4 + 2O2 ---> CO2 + 2H20

Then using dimension analysis:

$4.60*10^{-3} g CH4 ( \frac{moleCH4}{16 g CH4}) * ( \frac{2mole O2}{mole CH4}) * ( \frac{32 g O2}{mole O2} ) = 0.0184 g O_{2}$

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When sodium carbonate is dissolved in water, it doesn’t dissociate, or break up, completely. There is always some sodium carbona

dusya [7]

It's a weak base because weak bases don't dissociate completely. If it did, then it would be a strong base.

7 0

3 years ago

What quantity can be calculated for a solid compound, given only the formula of the compound and the periodic table of elements

gavmur [86]

When you have both of these ( The periodic table of elements and the formula of your compound) you are able to calculate the R.M.M ( Relative atomic mass) of that compound.
For example the formula of a NaCl ( Table salt ) has the elemnt Na and Cl.
We look at the atomic mass of both of these compounds
Na - 23
Cl - 35.5
R.M.M = 23 +35.5 = 58.5
Hope this helps :).

6 0

2 years ago

How does energy change when you eat food?

iris [78.8K]

It changes to chemical energy

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3 years ago

Balance CH2O + H2 > CH3OH

Tju [1.3M]

Answer:

I believe it is balanced

Explanation:

8 0

2 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol = 12.81 grams

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = 75.70 grams CO2

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = 34.29 grams

3 0

3 years ago