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Anestetic [448]
3 years ago
9

https://Suppose that the concentration of CO2 available for the Calvin cycle decreased by 50% (because the stomata closed to con

serve water). Which statement correctly describes how O2 production would be affected? (Assume that the light intensity does not change.)/105793539/ch-10-flash-cards/ . A. The rate of O2 production would remain the same because light reactions are independant of Calvin Cycle B. The rate of O2 production would remain the same because the light intensity did not change. C. The rate of O2 production would decrease because the rate of ADP
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer: C. The rate of O2 production would decrease because the rate of ADP and NADP+ production by the Calvin cycle would decrease

Explanation: It is true that the Calvin Cycle is not directly dependent on light, but it is indirectly dependent on light because the energy carriers (ATP and NADPH) are produced by light-dependent reactions. Carbon dioxide and two other components (ribulose bisphosphate carboxylase and three molecules of ribulose bisphosphate) are the ones to initiate the light-independent reactions. Thus, the concentration of CO2 will determine the rate of ADP and NADP+ production, which will directly affect the rate of production of O2.

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What's the molar mass of Li3p
spayn [35]

Answer:

51.79g Li₃P.

Explanation:

Li has a molar mass of 6.94 g (since there are 3, you multiply it 3 times) and P has a molar mass of 30.97 g. 6.94(3) + 30.97 = 51.79g.

4 0
3 years ago
A 50.0 mL sample of an aqueous H2SO4 solution is titrated with a 0.375 M NaOH solution. The equivalence point is reached with 62
vesna_86 [32]

Answer: The concentration of H_2SO_4 is 0.234 M

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity H_2SO_4 = 2

M_1 = molarity of H_2SO_4 solution = ?

V_1 = volume of  H_2SO_4 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 0.375 M

V_1 = volume of  NaOH solution = 62.5 ml

Putting in the values we get:

2\times M_1\times 50.0=1\times 0.375\times 62.5

M_1=0.234M

Therefore concentration of H_2SO_4 is 0.234 M

6 0
3 years ago
Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2
Mkey [24]

Answer:

ΔG°rxn = +50.8 kJ/mol

Explanation:

It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

ΔH°rxn = 136.5kJ/mol

And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

ΔH°rxn = 0.2875kJ/molK

And replacing in (1) at 298K:

ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

<em>ΔG°rxn = +50.8 kJ/mol</em>

<em />

7 0
3 years ago
Read 2 more answers
State and Explain the trend in the atomic radius down a group
balu736 [363]

Explanation:

<h2>The number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital.</h2>
7 0
1 year ago
Why does ice float after it crystallizes? Its mass is greater than water. Its density is greater than water. Its density is less
tekilochka [14]

Ice floats after it crystallizes because ITS DENSITY IS LESS THAN THAT OF WATER.

When a quantity of water is cools down by reducing its temperature, the molecules of the water lose kinetic energy and slow down in their movement. As the water is cooling down, it is volume is expanding. When the temperature reaches zero degree Celsius, the water becomes ice. At this point, the ice can float on water because its density is less than that of water; this is as a result of the spaces that now exist in the ice structure.

7 0
3 years ago
Read 2 more answers
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