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3 years ago

7. Calculate the amount of energy required heat 100.g to H2O(s) changes to H2O(l) at 0°C

Chemistry

1 answer:

Brrunno [24]3 years ago

8 0

Answer:

33300J

Explanation:

Given parameters:

Mass of ice = 100g

Unknown:

Amount of energy = ?

Solution:

This is a phase change process from solid to liquid. In this case, the latent heat of melting of ice is 3.33 x 10⁵ J/kg.

So;

H = mL

m is the mass

L is the latent heat of melting ice

Now, insert the parameters and solve;

H = mL

mass from gram to kilogram;

100g gives 0.1kg

H = 0.1 x 3.33 x 10⁵ = 33300J

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One sphere has a radions of 181 cm, another has a radius of 5.01 cm. What is the difference in volume (in cubic centimeters) bet

PSYCHO15rus [73]

Answer:

$2.4*10^{7}cm^{3}$

Explanation:

First you should calculate the volume of a big sphere,so:

$V_{big}=\frac{4}{3}\pi r^{3}$

$V_{big}=\frac{4}{3}\pi (181cm)^{3}$

$V_{big}=2.4*10^{7}cm^{3}$

Then you calculate the volume of a small spehre, so:

$V_{small}=\frac{4}{3}\pi r^{3}$

$V_{small}=\frac{4}{3}\pi (5.01cm)^{3}$

$V_{small}=5.3*10^{2}cm^{3}$

Finally you subtract the two quantities:

$V_{big}-V_{small}=2.4*10^{7}cm^{3}-5.3*10^{2}cm^{3}$

$V_{big}-V_{small}=2.4*10^{7}cm^{3}$

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3 years ago

b) 2C2H2(g) + 5O2(g)⟶4CO2(g) + 2H2O(l) 6.54 Calculate the heats of combustion for the following reactions from the standard enth

IRINA_888 [86]

Answer:

a. ΔH⁸ = -1420 kJ/mol b. ΔH⁸ = -1144.84 kJ/mol

Explanation:

C₂H₄ (g) + 3 O₂ (g) ------------------------ 4 CO₂ (g) + 2 H₂O (l) ΔH⁸ = ?

ΔH⁸f kJmol 52.47 0 -399.5 -285.83

ΔH⁸ = 2(-399.5) + 2 (-285.83) - (52.47)

ΔH⁸ = -1420 kJ/mol

H₂S (g) + 3 O₂ (g) ---------------------- 2 H₂O (l) + 2 SO₂ (g)

ΔH⁸f kJmol -20.50 0 -285.83 -296.84

ΔH⁸ = 2(-285.83) + 2 (-296.84) - (-20.50)

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4 0

3 years ago

Be sure to answer all parts.

MrRissso [65]

Answer: The molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

Explanation:

Molarity is the number of moles of a substance present in liter of a solution.

And, moles is the mass of a substance divided by its molar mass.

(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol$

Now, molarity of ethanol solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M$

(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol$

Now, molarity of sucrose solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M$

(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol$

Now, molarity of sodium chloride solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M$

Thus, we can conclude that the molarity of each of the given solutions is:

(a) 1.38 M

(b) 0.94 M

(c) 1.182 M

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