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3 years ago

PLEASE HELP DUE IN 20 MINS I NEED HELP RIGHT NOW

Chemistry

2 answers:

butalik [34]3 years ago

8 0

That looks like cells of a multicellular organism, so B.

CaHeK987 [17]3 years ago

8 0

The little compartments are mostly likely cells, so C and D can be eliminated. Since one must zoom in to see cells, the image is most likely not large enough to be able to identify the organism. Therefore, my best guess would be B.) Cells of a multicellular organism.

I hope this helped you come to a conclusion!

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Competition increases as _____.

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Is there unlimited space in space?

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4 years ago

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The reaction A + 2B → products was found to follow the rate law: rate = k[A] 2[B]. Predict by what factor the rate of reaction w

Alexus [3.1K]

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of A and the of the concentration of B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

[A] = 2x, [B] = 3y

The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

5 0

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

Read 2 more answers