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Rufina [12.5K]
3 years ago
14

HELP!! WILL GIVE BRAINLIEST!!! EASY!!

Chemistry
2 answers:
Leviafan [203]3 years ago
8 0

Answer:

D) Different substances than before are produced

Explanation:

In a chemical reaction, only the atoms present in the reactants can end up in the products. No new atoms are created, and no atoms are destroyed. In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

AysviL [449]3 years ago
4 0

Answer:

I'm almost positive its d

Explanation:

hope this helps <3

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A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
Bad White [126]

Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

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