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3 years ago

Which describes the role of energy in photosynthesis and cellular respiration?

Chemistry

2 answers:

r-ruslan [8.4K]3 years ago

8 0

Answer: A). Energy is needed to start photosynthesis and is a product of cellular respiration.

Explanation:

Energy in the form of sunlight is absorbed by the plants so as to prepare their food (carbohydrates)in the process of photosynthesis.

The energy is produced in the process of cellular respiration. The cellular respiration involves the break down of glucose molecules in the food in the form of energy currencies called as ATP (Adenosine triphosphate) molecules. This form of energy is required for regulating cellular metabolism and vital functions.

bagirrra123 [75]3 years ago

5 0

Answer:

A) Energy is needed to start photosynthesis and is a product of cellular respiration.

Explanation:

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The first excited vibrational energy level of diatomic chlo- rine (Cl2) is 558 cm^-1 above the ground state. Wave- numbers, the

Usimov [2.4K]

Answer:

The answer is "0.0000190 and 2.7 J".

Explanation:

$\to P_4=\frac{e^{-\beta}(4+\frac{1}{2}) hev}{2vb}\\\\$

$=\frac{e^{-9(1.35)}}{0.278v}\\\\=\frac{e^{-12.5}}{0.278}\\\\=\frac{0.000005285}{0.278}\\\\=0.0000190$

Given:

$h=6.626 \times 10^{-34}\\\\c=3\times 10^{10}\\\\v=558\ cm^{-1}\\\\K=1.38 \times 10^{-23}\\\\ T=298\\\\$

$E=\frac{hcv}{KT}\\\\$

by putting the value into the above formula so, the value is 2.7 J

6 0

3 years ago

Is anybody good in Chemistry that can help me with this?

sesenic [268]

Answer:

just replace the 9 mole with 3.68 g of Al .

I think it will help you.

5 0

3 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$

6 0

3 years ago

What 2 chemicals react together during photosynthesis

Nitella [24]

Answer:

The photosynthesis chemical equation states that the reactants (carbon dioxide, water and sunlight), yield two products, glucose and oxygen gas. The single chemical equation represents the overall process of photosynthesis.

Explanation:

Definition: the process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.

Hope this helps!

Can I get brainliest?

5 0

3 years ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

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