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3 years ago

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One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.00 liters against a constant external pressure

of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore significant figures for this problem. (T = 300 K; 1 L·atm = 101.3 J)

1 answer:

Leya [2.2K]3 years ago

4 0

Answer : The work done on the surroundings is, 709.1 Joules.

Explanation :

The formula used for isothermally irreversible expansion is :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure = 1.00 atm

$V_1$ = initial volume of gas = 1.00 L

$V_2$ = final volume of gas = 8.00 L

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

$w=-p_{ext}(V_2-V_1)$

$w=-(1.00atm)\times (8.00-1.00)L$

$w=-7L.atm=-7\times 101.3J=-709.1J$

The work done by the system on the surroundings are, 709.1 Joules. In this, the negative sign indicates the work is done by the system on the surroundings.

Therefore, the work done on the surroundings is, 709.1 Joules.

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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

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

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

3 years ago

What would the volume of gas be at 150 c if it had a volume of 693 ml at 45 c

marissa [1.9K]

Answer:

$\boxed{\text{922 mL}}$

Explanation:

The pressure is constant, so we can use Charles' Law to calculate the volume.

$\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}$

Data:

V₁ = 693 mL; T₁ = 45 °C

V₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert temperature to kelvins

T₁ = ( 45 + 273.15) = 318.15 K

T₂ = (150 + 273.15) = 423.15 K

(b) Calculate the volume

$\dfrac{ 693}{318.15} = \dfrac{ V_{2}}{423.15}\\\\2.178 = \dfrac{ V_{2}}{423.15}\\\\V_{2} = 2.178 \times 423.15 = \boxed{\textbf{922 mL}}$

5 0

3 years ago

One of the reactions in a blast furnace used to reduce iron is shown above. How many grams of Fe2O3 are required to produce 15.5

Salsk061 [2.6K]

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of $Fe_2O_3$ = 160 g/mole

First we have to calculate the moles of iron.

$\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles$

Now we have to calculate the moles of $Fe_2O_3$ .

The balanced reaction is,

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of $Fe_2O_3$

So, 0.276 moles of iron obtained from $\frac{0.276}{2}=0.138$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3$

$\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g$

Therefore, the amount of $Fe_2O_3$ required are, 22.1 grams.

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3 years ago

If you have 10 grams of Lithium Oxide what will the volume be? Show your work

gogolik [260]

Answer:

Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L

Explanation:

STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.

According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .

Molar mass of Lithium Oxide = 29.8 g/mol

$Li_{2}O$ = 2(6.9) + 15.99 = 29.8

Mass of 1 mole $Li_{2}O$ = 29.8 g

1 mole $Li_{2}O$ occupies, Volume = 22.4 L

29.8 g $Li_{2}O$ occupies, V = 22.4 L

1 g $Li_{2}O$ occupies ,V = $\frac{22.4}{29.8}$

1 g $Li_{2}O$ occupies ,V = 0.7516 L

10 g tex]Li_{2}O[/tex] occupies ,V = $0.7516 \times10$ L

V = 7.52 L

So, volume occupied by Lithium Oxide At STP is 7.52 L

4 0

3 years ago

The chemical reaction in which compounds break up into simpler Constituents is a.

coldgirl [10]

Decomposition its the answer

8 0

3 years ago

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