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vaieri [72.5K]
3 years ago
15

1. How many joules of heat are required to raise the temperature of 750 g of water from 11.0 oC to 19.0 oC?

Chemistry
1 answer:
Leya [2.2K]3 years ago
6 0

Answer:

  1. 25080 J
  2. 146.9 g
  3. 92.58 °C
  4. 0.808 J/g°C
  5. 117.09 g
  6. a. 1708.8 kJ  b.1246.56 kJ
  7. 368.55 kJ
  8. 6.81 kJ
  9. 5.50 grams of methane produces more heat than 5.5 grams of propane.

Explanation:

  1. The specific heat capacity of water=4.18 J/gK

The enthalpy change is calculated using the formula: ΔH=MC∅ where ΔH is the change in enthalpy, M the mass of the substance, C the specific heat capacity of the substance and ∅ the temperature change.

Thus, ΔH= 750g × 4.18 J/gK × (19-11)K

=25080 J

2. Enthalpy change= mass of substance × specific heat capacity of the substance× Change in temperature.

ΔH= MC∅

M= ΔH/(C∅)

Substituting for the values in the question.

M=8750 J/(0.9025/g°C×66.0 °C)

=146.9 grams

3. Enthalpy change =mass × specific heat capacity × Temperature

ΔH= MC∅

∅ = ΔH/(MC)

=6500 J/(250 g × 4.18 J/g°C)

=6.22° C

Final temperature =98.8 °C - 6.22°C

=92.58 °C

4. Specific heat capacity =mass × specific heat capacity × Temperature change.

ΔH=MC∅

C= ΔH/(M∅)

Substituting with the values in the question.

C = 4786 J/(89.0 g×(89.5° C-23°C))

=0.808 J/g°C

5. Heat lost lost copper is equal to the heat gained by water.

ΔH(copper)= ΔH(water)

MC∅(copper)=MC∅(water)

M×0.385 J/g°C× (75.6°C- (19.1 °C+5.5°C))=100.0g×4.18 J/g°C×5.5 °C

M=(100.0g×4.18J/g°C×5.5°C)/(0.385 J/g°C×51 °C)

=117.09 grams.

6 (a). From the equation 1 mole of methane gives out 890.4 kJ

There fore 2 moles give:

(2×890.4)/1= 1780.8 kJ  

(b) 22.4 g of methane.

Number of moles= mass/ RFM

RFM=12 + 4×1

=16

No. of moles =22.4 g/16g/mol

=1.4 moles

Therefore 1.4 moles produce:

1.4 moles × 890.4 kJ/mol=

=1246.56 kJ

7. From the equation, 2 moles of aluminium react with ammonium nitrate to produce 2030 kJ

Number of moles = mass/RAM

Therefore 9.75 grams = (9.75/26.982) moles of aluminium.

=0.3613 moles.

If 2 moles produce 2030 kJ, then 0.3613 moles produce:

(0.3631 moles×2030 kJ)/2

=368.55 kJ

8. From the equation, 4 moles of ammonia react with excess oxygen to produce 905.4 kJ of energy.

Number of moles= mass/molar mass

RMM= 14+3×1= 17

Therefore 0.5113 grams of ammonia = (0.5113 g/17g/mole) moles

= 0.0301 moles

If 4 moles produce 905.4 kJ, then 0.0301 moles produce:

(0.0301 moles×905.4 kJ)/4 moles

=6.81 kJ

9. From the equations, one mole of methane produces 890 kJ of energy while one mole of propane produces 2043 kJ.

Lets change 5.5 grams into moles of either alkane.

Number of moles= Mass/RMM

For propane, number of moles= 5.5g/ 44.097g/mol

=0.125 moles

For methane number of moles =5.5 g/ 16g/mol

=0.344 moles

0.125 moles of propane produce:

0.125 moles×2043 kJ/mol

=255.375kJ

0.344 moles of methane produce:

0.344 moles× 890 kJ/mol

= 306.16kJ

Therefore, 5.5 grams of methane produces more heat than 5.5 grams of propane.

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Answer:

(B) 0.160 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For K_2CO_3 :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of K_2CO_3 :

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles of K_2CO_3 = 0.004 moles

For Ba(NO_3)_2 :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of Ba(NO_3)_2 :

Moles=0.400\times {30.0\times 10^{-3}}\ moles

Moles of Ba(NO_3)_2  = 0.012 moles

According to the given reaction:

K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate  = 0.004 moles

Available moles of barium nitrate  =  0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles  (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

4 0
3 years ago
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Explanation:

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From the question above, when water was added to the sodium acetate in the flask, SOME of the chemical dissolved into the water, meaning that some remained undissolved. This is because a given volume of water can only dissolve a certain amount of chemical in it at room temperature. If more chemical is added to such a solution, the chemical will remain undissolved. Such a chemical solution is said to be a SATURATED SOLUTION.

A saturated solution of a solute at a particular temperature is on which contains as much solute as it can dissolve at that temperature in the presence of undissolved solute particles.

Unsaturated solution is a type of solution that dissolves all its solutes with no presence of undissolved solute.

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Answer:

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