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Masteriza [31]
2 years ago
11

How much energy (heat) is required to convert 52.0 g of ice at –10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C Spe

cific heat of water 4.18 J/g • °C Specific heat of steam 1.84 J/g • °C Molar heat of fusion 6.02 kJ/mol Molar heat of vaporization 40.7 kJ/mol
Chemistry
1 answer:
dexar [7]2 years ago
5 0

Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ

Explanation:

Using this formular, q = [mCpΔT] and = [nΔHfusion]

The energy that is needed in the different physical changes is thus:

The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:

q = [mCpΔT]

q = 52.0 x 2.09 x 10

q = 1.09 kJ

While from 0°C to 100°C is calculated as:

q = [mCpΔT]

q = 52.0 x 4.18 x 100

q = 21.74 kJ

And for fusion at 0°C is called Heat of fusion and would be given as:

q = n ΔHfusion

q = 52.0 / 18.02 x 6.02

q = 17.38 kJ

And that required for vaporization at 100°C is called Heat of vaporization and it's given as:

q = n ΔHvaporization

q = 52.0 / 18.02 x 40.7

q = 117.45 kJ

Add up all the energy gives 157.8 kJ

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Answer:

The heat of combustion is -25 kJ/g = -2700 kJ/mol.

Explanation:

According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.

Qcomb + Qcal = 0

Qcomb = - Qcal

The heat absorbed by the calorimeter can be calculated with the following expression.

Qcal = C × ΔT

where,

C is the heat capacity of the calorimeter

ΔT is the change in temperature

Then,

Qcomb = - Qcal

Qcomb = - C × ΔT

Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ

Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:

\frac{-5.0kJ}{0.1964g} =-25kJ/g

The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:

\frac{-25kJ}{g} .\frac{108g}{1mol} =-2700kJ/mol

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3 years ago
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It's B ------------------------------------

7 0
3 years ago
Read 2 more answers
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

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3 years ago
Rn-222 has a half-life of 3.82 days. If 25.0 g of Radon-22 was originally present, approximately how many grams would be left af
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Answer:

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Explanation:

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Let's round that up to 4

That means 15 days is around 4 half lives

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