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3 years ago

If a gas is measured to have a pressure of 1.2 atm at 227 K, and its origina

Chemistry

1 answer:

ICE Princess25 [194]3 years ago

7 0

Answer:

1.05

Explanation:

P1 over t1 is equal to P2 over T2, then solve for X

X/199= 1.2/227

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To how much water should 100 ml of 18m sulfuric acid be added to prepare a 1.5m solution

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Answer:

Take 100 ml of a 18 molar solution. The total number of moles is (1 liter/1000 ml) 100 ml 18 moles is 1.8 moles.

1.5 moles in 1 liter so If 1.1 liters of water is added, the total volume is 1.2 liters and 1.8 moles are dissolves in it. 1.8 moles/ 1.2 liters is 1.5 moles per liter.

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3 years ago

How to prepare super saturated solution of copper sulphate ? Explain the experiment

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Explanation:

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2 years ago

#1: Which of the following is an exothermic reaction?

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3 years ago

Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions.

sammy [17]

A electrochemical reaction is said to be spontaneous, if $E^{0} cell is positive.$

Answer 1:
Consider reaction: Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s)

The cell representation of above reaction is given by;
 $S^{2-}/S // Ni^{2+}/Ni$

Hence, $E^{0}cell = E^{0} Ni^{2+/Ni} - E^{0} S/S^{2-}$
we know that, ${E^{0} Ni^{2+}/Ni = -0.25 v$
and ${E^{0} S/ S^{2-} = -0.47 v$

Therefore, $E^{0} cell$ = - 0.25 - (-0.47) = 0.22 v

Since, $E^{0} cell$ is positive, hence cell reaction is spontaneous
.....................................................................................................................

Answer 2:
Consider reaction: Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

The cell representation of above reaction is given by;
 $H_{2} / H^{+} // Pb^{2+} /Pb$

Hence, $E^{0}cell = E^{0} Pb/Pb^{2+} - E^{0} H_{2}/H^{+}$
we know that, ${E^{0} Pb^{2+}/Pb = -0.126 v$
and ${E^{0} H_{2}/ H^{+} = -0 v$

Therefore, $E^{0} cell$ = - 0.126 - 0 = -0.126 v

Since, $E^{0} cell$ is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3:
Consider reaction: 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)

The cell representation of above reaction is given by;
 $Cr/Cr^{2+} // Ag^{+}/Ag$

Hence, $E^{0}cell = E^{0} Ag^{+}/Ag - E^{0} Cr/Cr^{2+}$
we know that, ${E^{0} Ag^{+}/Ag = -0.22 v$
and ${E^{0} Cr/ Cr^{2+} = -0.913 v$

Therefore, $E^{0} cell$ = - 0.22 - (-0.913) = 0.693 v

Since, $E^{0} cell$ is positive, hence cell reaction is spontaneous

8 0

4 years ago

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