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2 years ago

What is the minimum mass of 25°c water required to melt all 35.1 g of ice?

1 answer:

5 0

ΔH fusion of water is 6.0 kJ/mol
mass of ice = 35.1 g
Molecular weight of water = 18.015 g / mole
number of moles of water = 35.1 / 18.015 = 1.95 mole
amount of heat required to melt ice = moles * Δ H fusion = 1.95 x 6 =11.7 kJ
mass of water at 25°C required to get this energy to a temperature 0°C
q = m C ΔT
11700 J = m * 4.184 * (25°- 0°)
m = 111.85 g

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How many liters of Cl2 gas will you have if you are using 63 g of Na?

ELEN [110]

Answer:

You will have 19.9L of Cl2

Explanation:

We can solve this question using:

PV = nRT; V = nRT/P

Where V is the volume of the gas

n the moles of Cl2

R is gas constant = 0.082atmL/molK

T is 273.15K assuming STP conditions

P is 1atm at STP

The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:

63g * (1mol / 70.906g) = 0.8885 moles

Replacing:

V = 0.8885mol*0.082atmL/molK*273.15K/1atm

V = You will have 19.9L of Cl2

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3 years ago

What you think would happen if global warming caused a decrease in all clouds

marta [7]

Answer:

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Explanation:

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3 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

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3 years ago

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