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2 years ago

15

When calcium hydroxide is formed, heat is released. Calculate the amount of heat (in kJ) released when 2 moles of calcium hydrox

ide is formed.
CaO(s) + H2O(I) —> Ca(OH)2(s) + 65.2 kJ

A. 32.6 kJ
B. 65.2 kJ
C. 130.4 kJ
D. 4.18 kJ

1 answer:

babunello [35]2 years ago

3 0

Answer:

C

Explanation:

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3 years ago

dmitriy555 [2]

Answer:

C

Explanation:

We know that the molar mass of KOH is 56 gmol-1

Molar mass of Mg(OH)2 is 58.31 g/mol

Now; the balanced reaction equation is; MgCl2 + 2KOH → Mg(OH)2 + 2KCI

Hence;

2 moles of KOH yields 1 mole of Mg(OH)2

4 moles of KOH yields;

4 moles of KOH/1 mol * 1 mole Mg(OH)2 /2 moles KOH * 58.31 g Mg(OH)2 / 1 mole Mg(OH)2

So option c is correct

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2 years ago

How many liters are in a .00813M solution that contains 1.55 g of KBr?

Sloan [31]

Answer:

1.602 L (or) 1602 mL

Explanation:

Molarity is the amount of solute dissolved per unit volume of solution. It is expressed as,

Molarity = Moles / Volume of Solution ----- (1)

Rearranging above equation for volume,

Volume of solution = Moles / Molarity -------(2)

Data Given;

Molarity = 0.00813 mol.L⁻¹

Mass = 1.55 g

First calculate Moles for given mass as,

Moles = Mass / M.mass

Moles = 1.55 g / 119.002 g.mol⁻¹

Moles = 0.0130 mol

Now, putting value of Moles and Molarity in eq. 2,

Volume of solution = 0.0130 mol / 0.00813 mol.L⁻¹

Volume of solution = 1.60 L

or,

Volume of solution = 1602 mL

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3 years ago

Complete this equation for the dissociation of K3PO4(aq). Omit water from the equation because it is understood to be present.

Anna35 [415]

Balanced chemical reaction (dissociation): K₃PO₄(aq) → 3K⁺(aq) + PO₄³⁻(aq).
K₃PO₄ is potassium phosphate, <span>a water-soluble </span>ionic salt.
In water potassium phosphate, ionic compound, dissociates on positive potassium ion (cations) and negative phosphate ions (anions).
Potassium has positive charge (+1), compound has neutral charge.

3 0

3 years ago

Read 2 more answers

A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane

son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

= 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

= 75 - 67.5

= 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

= 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

= 25 - 21.25

= 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

= 64.125 moles

$CO$ formed is = 0.05 x 67.5

= 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

= 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

= 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21}$

= 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

= 2 x 64.125

= 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

= 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3 = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

= 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.

6 0

3 years ago