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Pachacha [2.7K]
3 years ago
12

In the combustion of ethane, shown above, how many moles of carbon dioxide can be produced from 1.00 mole of ethane? A. 0.500 mo

le B. 1.00 mole C. 2.00 moles D. 4.00 moles
Chemistry
1 answer:
Daniel [21]3 years ago
6 0
The balanced chemical equation for the <span>combustion of ethane is 
2C</span>₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)

The stoichiometric ratio between C₂H₆(g) and CO₂(g)  is 1 : 2

Hence,
    moles of 
CO₂(g) produced = moles of reacted C₂H₆(g) x 2
                                                 
= 1.00 mol x 2
                                                  = 2.00 mol

Hence, the correct answer is "C".
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Answer:

Models 1 and 2 would be in the same group.

Explanation:

They have the same amount of valence electrons. Valence electrons are electrons in the outermost electron cloud. This means they have the same properties, therefore are in the same Group on the periodic table.

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4 years ago
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Which of these is a physical property?
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Answer:

malleable

Explanation:

able to be bended into shape

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3 years ago
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Determine the molar mass of an unknown monoprotic acid to two decimal places if 20.01 ml of a 0.098 m naoh solution were used to
natita [175]
By using the following formula we can calculate the volume;
C₁V₁ = C₂V₂
C₁ = concentration of starting solution = 0.098m
V₁ = volume of starting solution that is needed for titration = 20.01ml
C₂ = Desired concentration of final solution = ?
V₂ = desired volume of final solution = 0.215g  = 0.215ml
Now putting the values in the formula;
0.098 x 20.01 = C₂ x 0.215
C₂ = 0.215 / 1.96098 = 0.109m = 0.11m
Thus, the answer is 0.11m.
4 0
3 years ago
If you want to prepare 2.00 L of a 0.800 M NaNO_3 solution from a NaNO_3 stock solution that is 1.5 M in concentration, how many
Leya [2.2K]

1066.7 ml of the stock solution must start with, preparing 2.00 L of a 0.800 M NaNO3 solution.

<h3>What is volume?</h3>

Volume is the space occupied by a three-dimensional object.

By the formula \rm M_1V_1= M_2V_2

M1 is the initial molarity, 1.5 M

M2 is the final molarity, 0.800 M

V1 is the initial volume

V2 is the final volume, 2.00 L

Putting the values in the equation

\rm 1.5\;M \times V_1= 0.800\;M \times 2.00\;L\\\\V_1 = \dfrac{ 0.800\;M \times 2.00\;L}{1.5\;M} = 1066.7\;ml

Thus, the initial volume is 1066.7 ml

Learn more about volume

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3 0
3 years ago
If there are 160 Calories in 28 g in 1 serving of Flamin' Hot Cheetos, how many gram(s) of Cheetos must you burn to raise the te
vagabundo [1.1K]

Answer:

3.5 g

Explanation:

The density of water is 1 g/mL. The mass (m) corresponding to 20.0 mL is 20.0 g.

We can calculate the heat (Q) required to raise the temperature of 20.0 mL of water 1 °C (ΔT).

Q = c × m × ΔT = 1 cal/g.°C × 20.0 g × 1 °C = 20 cal

where,

c is the specific heat capacity of water

There are 160 calories in 28 g of Cheetos. The mass that releases 20 cal is:

20 cal × (28 g/160 cal) = 3.5 g

6 0
4 years ago
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