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3 years ago

8

Rewrite each equation below with the delta h value included with either the reactants or he products , and identify the reaction

as endothermic or exothermic

1 answer:

grin007 [14]3 years ago

5 0

I will take a stab at it, but there are not equations, did you forget them?

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A gas has an initial volume of 15 L. If the temperature increases from 330 K to 450 K, what is the new volume.

Artist 52 [7]

Answer:

20.(45)L or about 20.4545L

Explanation:

PV = nRT

Where:

P - pressure

V - volume

n - number of particle moles

R - a constant

T - temperature in K

We can assume the P and n (and definitely R) stay the same, so we infer that

$V_1 = \frac{nRT_1}{P} = 15L\\V_2 = \frac{nRT_2}{P}\\V_1 / V_2 = \frac{nRT_1}{P} / \frac{nRT_2}{P} = \frac{T_1}{T_2}\\\\15L / V_2 = \frac{330K}{450K} = \frac{11}{15}\\\\V_2 = 15L \cdot \frac{15}{11} = 20.(45)L$

4 0

2 years ago

1) The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K.

Morgarella [4.7K]

Answer:

The enthalpy of the reaction is 64.9 kJ/mol.

Explanation:

$H_2 + \frac{1}{2}O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ$ ..[1]

$CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ$ ..[2]

$C_2H_6 + \frac{7}{2}O_2\rightarrow 2CO_2 + 3H_2O,\Delta H_3= -1559.9 kJ$ ..[3]

$2CH_4(g)\rightarrow C_2H_6(g) + H_2(g),\Delta H_4=?$ ..[4]

2 × [2] - [1]- [3] = [4] (Using Hess's law)

$\Delta H_4=2\times \Delta H_2 -\Delta H_1 -\Delta H_3$

$\Delta H_4=2\times (-890.4 kJ)-(-285.8 kJ) -(-1559.9 kJ)$

$\Delta H_4=64.9 kJ/mol$

The enthalpy of the reaction is 64.9 kJ/mol.

3 0

2 years ago

!!!!HELP PLEASE!!!!!Which of the following molecules has a trigonal pyramidal shape?

aksik [14]

Answer:

No.D is the molecules that has trigonal pyramidal sape

8 0

3 years ago

Read 2 more answers

3.) All matter has both physical and chemical properties. A physical property is one that does not change the chemical nature of

dexar [7]

Answer:

D

Explanation:

It would be D because you are observing the reaction and don’t change anything

8 0

2 years ago

What is the molality of an aqueous solution containing FeCl3 (MM = 162.2 g/mol) with a mole fraction of FeCl3 of 0.15?

Natalka [10]

Answer:

10 m

Explanation:

The mole fraction of FeCl₃ of 0.15, that is, per mole of solution, there are 0.15 moles of FeCl₃ and 1 - 0.15 = 0.85 moles of water.

The molar mass of water is 18.02 g/mol. The mass corresponding to 0.85 moles is:

0.85 mol × 18.02 g/mol = 15 g = 0.015 kg

The molality of FeCl₃ is:

m = moles of solute / kilogram of solvent

m = 0.15 mol / 0.015 kg

m = 10 m

3 0

3 years ago