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nasty-shy [4]
3 years ago
14

What mass of carbon dioxide is produced upon complete combustion

Chemistry
2 answers:
SOVA2 [1]3 years ago
8 0

Answer:

1 mole CO2 = 44g i.e.by the complete combustion of 12g of carbon, 44g of CO2 is produced.

Explanation:

lianna [129]3 years ago
4 0

Answer:

1 mole CO2 = 44g i.e.by the complete combustion of 12g of carbon, 44g of CO2 is produced.

Explanation:

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larisa86 [58]

Answer: closed system.

Explanation:

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I need to know the answer to that
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im not sure which one to answer, and i can hardly see the text.

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Why is it better to give the actual speed of an object rather than to simply say it moved fast or it moved slowly
DIA [1.3K]
Specificity. It’s really loose to say that something is fast, since speed can be scalarly linked and relative. I could say that both a car on the highway is fast, but so is the speed of light. The actual speed of something helps to do away with the arbitrary nature of using “fast” and “slow”; however, we’re still at step one of the person who is receiving the information is unfamiliar with the scale that the actual speed is defined in.
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3 years ago
A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.
Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K

Putting values in above equation, we get:

\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L

Therefore, the new volume of the air is, 6.83 L

6 0
4 years ago
Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
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