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3241004551 [841]
4 years ago
6

I'm looking at the barometer in my lab as I type this question. It says the pressure is 968 torr. What is this pressure expresse

d in atm
Chemistry
1 answer:
Ksenya-84 [330]4 years ago
7 0

Answer:

1.27 atm.

Explanation:

The following data were obtained from the question:

Pressure (P) in torr = 968 torr

Pressure (P) in atm =...?

Thus, we can convert from torr to atm by doing the following:

Recall:

760 torr = 1 atm

Therefore,

968 torr = 968/760 = 1.27 atm

Therefore, 968 torr is equivalent to 1.27 atm.

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Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, from the following data: 2C6H6(l) + 15O2(g)→12CO2(g)+6H2O(l) ΔH∘=−653
Pachacha [2.7K]

Answer: 48.6 kJ/mol

Explanation:

The balanced chemical reaction is,

2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_6H_6}\times \Delta H_{C_6H_6})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-6534.0=[(12\times -393.5)+(6\times -285.8)]-[(15\times 0)+(2\times \Delta H_{C_6H_6})]

\Delta H_{C_6H_6}=48.6kJ/mol

Therefore, the enthalpy change for benzene is 48.6 kJ/mol

4 0
3 years ago
How many grams of 48.0 wt% NaOH (FM 40.00) should be diluted to 1.00 L to make 0.11 M NaOH? (Enter your answer using two digits.
kifflom [539]

Answer:

The answer is 916.67 g

Explanation:

48.0 wt% NaOH means that there are 48 g of NaOH in 100 g of solution. With this information and the molecular weight of NaOH (40 g/mol), we can calculate the number of mol there are in 100 g of this solution:

\frac{48 g NaOH}{100 g solution} x \frac{1 mol NaOH}{40 g} = 0.012 mol NaOH/100 g solution

Finally, we need 0.11 mol in 1 liter of solution to obtain a 0.11 M NaOH solution.

0.012 mol NaOH ------------ 100 g solution

0.11 mol NaOH------------------------- X

X= 0.11 mol NaOH x 100 g/ 0.012 mol NaOH= 916.67 g

We have to weigh 916.67 g of 48.0%wt NaOH and dilute it in a final volume of 1 L of water to obtain a 0.11 M NaOH solution.

7 0
3 years ago
When dinitrogen pentaoxide, a white solid, is heated, it decomposes to produce nitrogen dioxide gas and
AysviL [449]

9.5314 L is the volume of nitrogen dioxide formed at 103.25 kPa and 22.75 °C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given data:

Oxygen produced - 1.618 gram

Decomposition of N_2O_5 takes place.

Find - Amount of NO_2 produced.

The decomposition reaction is as follows -

2N_2O_5--> 4NO_2 + O_2

Moles of O_2 gas =\frac{1.6}{16}  =0.1 moles.

1 mole of O_2 is produced from 2 moles of dinitrogen pentoxide

0.1 mole of O_2  will be produced from = 0.2 moles.

Now, 2 moles of dinitrogen pentoxide produce 4 moles of NO_2

NO_2 produced will be - 0.4 moles.

Weight of NO_2 produced - 0.4 X 46

Weight of NO_2  produced - 18.4 gram

Thus, grams of NO_2 produced are 18.4

Now calculate the volume of NO_2

Given data are:

P=103.25 kPa =1.01899827 atm

T= 22.75 °C +273 = 295.75 K

n=0.4 moles

V=?

R= 0.0821 liter·atm/mol·K

Putting the value in PV=nRT

V =  \frac{nRT}{P}

V =  \frac{0.4 \;moles \;X \;0.0821\; liter\;atm/\;mol \;K X \;295.75 \;K}{1.01899827 atm}

V= 9.5314 L

Hence, 9.5314 L is the volume of nitrogen dioxide formed at 103.25 kPa and 22.75 °C.

Learn more about the ideal gas equation here:

brainly.com/question/13450124

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8 0
2 years ago
Predict the organic product formed when bzcl reacts with isopropyl alcohol. bzcl = benzoyl chloride.
Levart [38]
Its Esterification reaction. Benzoyl chloride on reaction with alcohols give corresponding ester. In this case the ester formed is as follow,

4 0
3 years ago
Any body tell me Iron salt and water produced what ​
kolezko [41]

iron salts react with water to form hydrated iron ( III ) oxide which is basically rust

hope that helps :)

7 0
3 years ago
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