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Nuetrik [128]
3 years ago
7

When 8.21 l of c3h8 (g) burn in oxygen, how many liters of oxygen are consumed? all gas volumes are measured at the same tempera

ture and pressure but not at stp?
Chemistry
1 answer:
just olya [345]3 years ago
5 0
<span>8.21 L of C3H8(g) Lets take c as the molar volume at that temperature.
    c L <><> 5c L
  C3H8 (g) + 5O2 (g) --> 3CO2 + 4H2O + Q
  8.21 L <><> x L
    x = (8.21 * 5c)/c = 8.21 * 5 = 41.05 L O2 consumed for a 100% yield.</span>
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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.

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3 years ago
Proszę napisać wzór kwasu karboksylowego o 7 atomach węgla sumaryczny i strukturalny.
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Answer:

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Carboxylic acids is a homologous series in which the compounds contain a functional group called the carboxyl group (-COOH). The general molecular formula for carboxylic acids is CnH2n+1COOH. Carboxylic acids contain at least one carboxyl group.

Or in Polish…

Kwasy karboksylowe to homologiczna seria, w której związki zawierają grupę funkcyjną zwaną grupą karboksylową (-COOH). Ogólny wzór cząsteczkowy dla kwasów karboksylowych to CnH2n + 1COOH. Kwasy karboksylowe zawierają co najmniej jedną grupę karboksylową.

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Further Explanation:

We can convert from number of representative particles to moles using the formula:

\boxed {no. \ of \ moles \ = \ ( given \ no. \ of \ particles) \ (\frac{1 \ mole}{\ 6.022 \ x 10^{23} particles})}

For this problem, we can calculate the number of moles by plugging in the given values to the equation above,

no. \ of \ moles \ = (5.3 \ x \ 10^{24} \ formula \ units \ AgNO_{3}) \ (\frac{1 \ mole \ AgNO_{3}}{6.022 \ x 10^{23} \ formula \ units AgNO_{3}}) \\\\\boxed {no. \ of moles \ AgNO_{3} \ = \ 8.8 \ moles}

Learn More

  1. Learn more about representative particles brainly.com/question/8969313
  2. Learn more about Avogadro's number brainly.com/question/229300
  3. Learn more about mole conversions brainly.com/question/1370888

Keywords: moles conversion, Avogadro's number

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