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3 years ago

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The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p

ressure. 1. q = 6020 j/mol

1 answer:

zysi [14]3 years ago

3 0

Answer : q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

$q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})$ (in terms of mass)

or, $q=\Delta H_{fusion}\times Moles$ (in terms of moles)

Now we have to calculate the value of q.

$q=6020J/mole\times 1Mole=6020J$

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of $\Delta e=0$

Now we have to calculate the value of w.

Formula used : $\Delta e=q+w$

where, q is heat required, w is work done and $\Delta e$ is internal energy.

Now put all the given values in above formula, we get

$0=6020J+w$

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

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Answer:

$V_{HCl}=5.05x10^{-3}L$

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$2*n_{HCl}=n_{Ba(OH)_2}$

In such a way, in terms of molarities and volumes, we can compute the required volume of hydrochloric acid as shown below:

$2*M_{HCl}V_{HCl}=M_{Ba(OH)_2}V_{Ba(OH)_2}\\\\V_{HCl}=\frac{M_{Ba(OH)_2}V_{Ba(OH)_2}}{2M_{HCl}} =\frac{0.101M*0.050L}{2*0.50M} \\\\V_{HCl}=5.05x10^{-3}L$

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3 years ago

Read 2 more answers

How many grams of water are produced when 4.50 L of

MA_775_DIABLO [31]

The answer for the following problem has been mentioned below.

<em><u>Therefore the mass of the water is 5.802 grams.</u></em>

Explanation:

Given:

volume of oxygen (V) = 4.50 L

Temperature (T) = 425 K

pressure of oxygen (P) = 2.50 atm

Gram molecular mass of oxygen (M) = 16.0 grams

To calculate:

mass of water (m)

We know;

According to the ideal gas equation;

P × V = n × R × T

As we know;

no of moles = $\frac{m}{M}$

m represents the mass of oxygen (m)

M represents the Gram molecular mass (M)

According to above mentioned equation;

P × V = n × R × T

P represents the pressure of the oxygen

V represents the volume of the oxygen

n represents the no of moles of the oxygen

R represents the universal gas constant

where,

the value of R is 0.0821 L atm/K moles

Substituting the values in the above equation;

2.50 × 4.50 = $\frac{m}{16.0}$ × 0.0821 × 425

11.25 = $\frac{m}{16.0}$ × 34.8925

180 = m × 34.8925

m = $\frac{180}{34.8925}$

m = 5.158 grams

Therefore the mass of the of oxygen is 5.158 grams

Now;

As we know;

$\frac{m_{1} }{M_{1} } = \frac{m_{2} }{M_{2} }$

where;

$m_{1}$ represents the mass of the oxygen

$M_{1}$ represents the gram molecular mass of the oxygen

$m_{2}$ represents the mass of the water

$M_{2}$ represents the gram molecular mass of water

From the above given formula,

$\frac{5.158}{16.0}$ = $\frac{m_{2} }{18}$

where;

Gram molecular weight of water = 18.0 u

$m_{2}$ = 5.802 grams

<em><u>Therefore the mass of the water is 5.802 grams.</u></em>

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2 years ago

In liquids, the attractive intermolecular forces are ________. In liquids, the attractive intermolecular forces are ________. st

Fantom [35]

Answer:

strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other.

Explanation:

In liquids, the attractive intermolecular forces are <u>strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other</u>.

Intermolecular forces are the forces of repulsion or attraction.

Intermolecular forces lie between atoms, molecules, or ions. Intramolecular forces are strong in comparison to these forces.

<u />

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2 years ago

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grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

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$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
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$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

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Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

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$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

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$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

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3 years ago

Phosphorus atomic radius is smaller than magnesium atomic radius <br> True or false

stealth61 [152]

Answer:

True

Explanation:

Atomic radius can be defined as a measure of the size (distance) of the atom of a chemical element such as hydrogen, oxygen, carbon, nitrogen etc, typically from the nucleus to the valence electrons. The atomic radius of a chemical element decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon). Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).

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2 years ago