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2 years ago

What is the chemical equation( formula ) of the reaction of photosynthesis

Chemistry

1 answer:

LekaFEV [45]2 years ago

4 0

Answer:

The process of photosynthesis is commonly written as: 6CO2 + 6H2O → C6H12O6 + 6O2.

Explanation:

This means that the reactants, six carbon dioxide molecules and six water molecules, are converted by light energy captured by chlorophyll (implied by the arrow) into a sugar molecule and six oxygen molecules, the products.

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For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the

Radda [10]

Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).
k₁ = 0,000643 1/s.
k₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

1/T₁ = 1/622 K = 0,0016 1/K.
1/T₂ = 1/666 K = 0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.

6 0

2 years ago

During nuclear fission and fusion, what is matter that seems to disappear actually converted into?

hjlf

Answer:

During nuclear fission and fusion matter that seems to disappear but is actually converted into energy. The amount of energy (E) produced in such a reaction can be calculated using Einstein's formula for the equivalence of mass and energy: E = mc^2.

Explanation:

4 0

2 years ago

Read 2 more answers

How is biodiversity related to niches?

Alja [10]

Answer:

C. The more niches in an ecosystem, the greater the biodiversity.

Explanation:

3 0

2 years ago

Read 2 more answers

A sample of (NH4)2SO4 contains 0.750 mole. what is the mass of the sample

steposvetlana [31]

The answer is 99.10.

5 0

1 year ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago