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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.

You might be interested in
How many moles of calcium chloride are needed to react completely with 6.2 moles of silver nitrate
Harlamova29_29 [7]

Answer:

3.1 mol

Step-by-step explanation:

The balanced equation is

CaCl₂ + 2AgNO₃ ⟶ Ca(NO₃)₂ + 2AgCl

You want to convert moles of AgNO₃ to moles of CaCl₂.

The molar ratio is 2 mol AgNO₃:1 mol CaCl₂

Moles of CaCl₂ = 6.2 mol AgNO₃ × (1 mol CaCl₂/2 mol AgNO₃)

Moles of CaCl₂ = 3.1 mol CaCl₂

The reaction will require 3.1 mol of CaCl₂.

8 0
3 years ago
Use the following equation to answer the questions and please show all work.
DIA [1.3K]

Answer:

a.36 g of water is produced.

b.64 g of O_{2} is consumed.

Explanation:

The reaction is 2H_{2} + O_{2}⇒2H_{2}O

a.

Given,

Weight of H_{2} reacted = 4g

Weight of 1 mole of H_{2} = 2\times1 = 2g

Therefore no. of moles of H_{2} reacted = \frac{4}{2} = 2 moles;

Also given,

Weight of O_{2} reacted = 32 g

Weight of 1 mole of O_{2} = 2\times16 = 32 g

Therefore no. of moles of O_{2} reacted = \frac{32}{32} = 1

We know that 2 moles of Hydrogen reacts with 1 mole of Oxygen to give 2 moles of water,

As we took 2 moles of Hydrogen and 1 mole of Oxygen,

Directly,from the equation we can tell 2moles of water will be produced.

Therefore no. of moles of H_{2} O produced = 2

Weight of 1 mole of water = 2\times 1+16 = 18

Therefore weight of H_{2}O produced = 2\times 18 = 36gm

b.

Given ,

72 g of H_{2}O is produced.

So,

no. of moles of H_{2}O produced =\frac{72}{18} = 4 moles

From equation For every 2 moles of water formed , 1 mole of oxygen must be required.

So for producing 4 moles of water,

No. of moles of Oxygen required = 2 moles.

Therefore weight of O_{2} reacted = 2\times32 = 64 g

Method 2:

Given,

8 g of H_{2} has reacted.

So,

no. of moles of H_{2} reacted = \frac{8}{2} = 4 moles.

From equation , we know that For every 2 moles of H_{2} reacted,1 mole of O_{2} will react.

Therefore,

No. of moles of O_{2} that reacts with 4 moles of H_{2} = 2\times1 = 2 moles

Therefore the weight of O_{2} reacted = 2\times 32 = 64 g

6 0
3 years ago
Breathing air that contains 4.0% by volume CO2 over time causes rapid breathing, throbbing headache, and nausea, among other sym
7nadin3 [17]

Answer:

1 mole of any ideal gas occupies the same volume as one mole of any other ideal gas under the same conditions of temperature and pressure.So 1 L CO2 has same number of moles as 1 L O2 and 1 L N2 etc.This means that, assuming all the gases in air are ideal gases, if CO2 is 4.0 % by volume then it is also 4.0 % by moles, because a volume of each gas has the same number of moles.

Explanation:

3 0
3 years ago
Urgent!!!
Yakvenalex [24]

The specific heat of the metal, given the data from the question is 0.60 J/gºC

<h3>Data obtained from the question </h3>

The following data were obtained from the question:

  • Mass of metal (M) = 74 g
  • Temperature of metal (T) = 94 °C
  • Mass of water (Mᵥᵥ) = 120 g
  • Temperature of water (Tᵥᵥ) = 26.5 °C
  • Equilibrium temperature (Tₑ) = 32 °C
  • Specific heat capacity of the water (Cᵥᵥ) = 4.184 J/gºC
  • Specific heat capacity of metal (C) =?

<h3>How to determine the specific heat capacity of the metal</h3>

The specific heat capacity of the sample of gold can be obtained as follow:

According to the law of conservation of energy, we have:

Heat loss = Heat gain

MC(T –Tₑ) = MᵥᵥC(Tₑ – Tᵥᵥ)

74 × C(94 – 32) = 120 × 4.184 (32 – 26.5)

C × 4588 = 2761.44

Divide both side by 4588

C = 2761.44 / 4588

C = 0.60 J/gºC

Thus, the specific heat capacity of the metal is 0.60 J/gºC

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

6 0
1 year ago
The actual yield of a product in a reaction was measured as 4.20 g. If the theoretical yield of the product for the reaction is
Norma-Jean [14]

\frac{actual yield}{theoretical yield}*100

\frac{4.20}{4.88}*100

<u>86.1%</u>

5 0
4 years ago
Read 2 more answers
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