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3 years ago

How many calories are required to raise the temperature of 105 g of water from 30.0°C to 70.0°C?

1 answer:

5 0

There number of calories that are required to raise the temperature of 105 g of water from 30.0 c to 70.0 c is calculated as follows

Q(heat)= mass(m) x specific heat capacity of water ( 1 cal/g/c) x delta T ( change in temperature

delta T = 70.0- 30.0= 40.0 c

Q= 105 g x 1 cal/g/c x 40 c = 4200 calories is needed to raise temperature of water from 30 c to 70 c

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Lisa [10]

Answer:

A) a chain reaction that is caused by nuclear fission

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3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

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3 years ago

Which of the following indicates the cathode rays has a large charge to mass ratio?

Anna007 [38]

Answer:

Gggg

Explanation:

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2 years ago

Which of these is an acid?

julsineya [31]

My best guess would be c

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3 years ago

Match the salt with the acid and base used to form it in a neutralizing reaction. K2SO4

emmasim [6.3K]

Answer:

KOH and H₂SO₄

Explanation:

Neutralization reaction:

It is the reaction in which acid and base react with each other and produce salt and water.

For example:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

1. Potassium hydroxide and sulfuric acid react to produce potassium sulfate salt and water.

2. Potassium hydroxide and phosphoric acid react to produce potassium phosphate and water.

H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O

3. Phosphoric acid sodium hydroxide react to produce sodium phosphate and water.

H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O

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3 years ago

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