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Sedbober [7]
2 years ago
9

Addition of 6 m hcl to which substance will not result in gas evolution?

Chemistry
1 answer:
saul85 [17]2 years ago
6 0

There is no gaseous product when NaNO3 reacts with 6M HCl.

We must note that gas evolution is one of the signs that a chemical reaction has taken place. Hence, we must consider the options and look out for an option in which reaction with HCl does not lead to a gaseous product. This is the option that does not lead to evolution of a gas.

Let us consider the reaction; NaNO3(s) + HCl(aq) ---> NaCl(s) + HNO3(aq), there is no gaseous product hence NaNO3 does not result in gas evolution upon addition of 6M HCl.

Missing parts;

Addition of 6 M HCl to which substance will NOT result in gas evolution?

(A) Al

(B) Zn

(C) K2CO3

(D) NaNO3

Learn more about evolution of gas: brainly.com/question/2186340

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This is because gas particles are free to move as they are not held in place by strong molecular forces while particles in a solid are

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How many miles can you drive, if your car gets 21.3 miles per gallon and you have 12.0 gallons of gas?
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255.6

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Read 2 more answers
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}

4 0
3 years ago
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