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valentinak56 [21]
3 years ago
14

6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)​

Chemistry
1 answer:
scoray [572]3 years ago
8 0

Taking into account the definition of calorimetry, 0.0185 moles of water are required.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

<h3>Mass of water required</h3>

In this case, you know:

  • Heat= 92.048 kJ
  • Mass of water = ?
  • Initial temperature of water= 34 ºC
  • Final temperature of water= 100 ºC
  • Specific heat of water = 4.186 \frac{J}{gC}

Replacing in the expression to calculate heat exchanges:

92.048 kJ = 4.186 \frac{J}{gC}× m× (100 °C -34 °C)

92.048 kJ = 4.186 \frac{J}{gC}× m× 66 °C

m= 92.048 kJ ÷ (4.186 \frac{J}{gC}× 66 °C)

<u><em>m= 0.333 grams</em></u>

<h3>Moles of water required</h3>

Being the molar mass of water 18 \frac{g}{mole}, that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as:

amount of moles=0.333 gramsx\frac{1 mole}{18 grams}

<u><em>amount of moles= 0.0185 moles</em></u>

Finally, 0.0185 moles of water are required.

Learn more about calorimetry:

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brainly.com/question/14057615?referrer=searchResults

brainly.com/question/24988785?referrer=searchResults

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What is the molarity of an aqueous solution that contains 78g of C6H12O6 dissolved in 2500 mL of solution?
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Answer:

\boxed {\boxed {\sf molarity = 0.17 \ M \ C_6H_12O_6}}

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

molarity = \frac {moles}{liters}

We are given grams of a compound and milliliters of solution, so we must make 2 conversions.

1. Gram to Moles

We must use the molar mass. First, use the Periodic Table to find the molar masses of the individual elements.

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  • H: 1.008 g/mol
  • O: 15.999 g/mol

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6(12.011)+12(1.008)+6(15.999)=180.156 g/mol

Use this number as a ratio.

\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}

Multiply by the given number of grams.

78 \ g \ C_6H_12O_6 *\frac {180.156 \ g\ C_6H_12 O_6}{ 1 \ mol \ C_6H_12O_6}

Flip the fraction and divide.

78 \ g \ C_6H_12O_6 *\frac { 1 \ mol \ C_6H_12O_6}{180.156 \ g\ C_6H_12 O_6}

\frac { 78 \ mol \ C_6H_12O_6}{180.156 }= 0.432958102977 \ mol \ C_6H_12O_6

2. Milliliters to Liters

There are 1000 milliliters in 1 liter.

\frac {1 \ L }{ 1000 \ mL}

Multiply by 2500 mL.

2500 \ mL* \frac {1 \ L }{ 1000 \ mL}

2500 * \frac {1 \ L }{ 1000 }= 2.5 \ L

3. Calculate Molarity

Finally, divide the moles by the liters.

molarity = \frac {0.432958102977 \ mol \ C_6H_12O_6}{ 2.5 \ L}

molarity = 0.173183241191 \ mol \ C_6H_12O_6/L

The original measurement has 2 significant figures, so our answer must have the same. That is the hundredth place and the 3 tells us to leave the 7.

molarity \approx 0.17 \ mol \ C_6H_12O_6 /L

1 mole per liter is also equal to 1 M.

molarity = 0.17 \ M \ C_6H_12O_6

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