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2 years ago

14

What happens to Hydrogen molecules after the light reaction in photosynthesis? Hydrogen is released into the atmosphere Hydrogen

bonds with Carbon Dioxide to form glucose Hydrogen bonds with Oxygen to form water.

1 answer:

vaieri [72.5K]2 years ago

5 0

Photosyntheis is a biochemical reaction that combines chemicals with biological species to produce energy. Atoms of hydrogen bond with carbon to form glucose.

<h3>What are the reactants and the products of the light reaction?</h3>

Carbon dioxide and hydrogen are the reactants that are used from the environment by the plants to make energy. Glucose and oxygen are the products.

The hydrogen atoms from the environment are absorbed by the plants and are combined with the carbon dioxide to yield glucose or sugar molecules for energy.

Therefore, option B. hydrogen and carbon dioxide react to produce glucose.

Learn more about photosynthesis here:

brainly.com/question/1388366

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Which of the following atoms will form an ion with the most negative charge?

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Answer:

sulfur will form an ion with negative charge

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2 years ago

(HELP I HAVE 10 MINS TO TURN IN) why is the nervous system in the human body a system?

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3 years ago

Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)

Sedbober [7]

Answer:

The yield of the product in gram is $\mathsf{{w_P}=0.26 \ gram}$

Explanation:

Given that:

the molecular mass weight of the product = 96.2 g/mol

the molecular mass of the reagent (3S)-2,2,-dibromo-3,4-dimethylpentane is 257.997 g

given that the millimoles of the reagent = 2,7 millimoles = $2.7 \times 10^{-3} \ moles$

We know that:

Number of moles = mass/molar mass

Then:

$2.7 \times 10^{-3} = \dfrac{ mass}{257.997}$

$mass = 2.7 \times 10^{-3} \times 257.997$

mass = 0.697

Theoretical yield = (number of moles of the product/ number of moles of reactant) × 100

i.e

Theoretical yield = $\dfrac{n_P}{n_R}\times 100\%$

where;

$n_P = \dfrac{w_P}{m_P}$ and $n_R = \dfrac{w_R}{m_R}$

Theoretical yield = $\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

Given that the theoretical yield = 100%

Then:

$100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%$

$\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}$

${w_P}=\dfrac{w_R \times m_P}{m_R}$

where,

$w_P$ = derived weight of the product

$m_P =$ the molecular mass of the derived product

$m_R =$ the molecular mass of the reagent

$w_R$ = weight in a gram of the reagent

${w_P}=\dfrac{w_R \times m_P}{m_R}$

${w_P}=\dfrac{0.697 \times 96.2}{257.997}$

$\mathsf{{w_P}=0.26 \ gram}$

8 0

3 years ago

14) How many molecules are there in 9.34 kg of water?

madreJ [45]

Answer:

Water has 18 grams per mole so 9 grams is 0.5 moles of water. There are 6.022* 10^23 molecules in a mole. Each water molecule contains one oxygen and two hydrogens.

8 0

3 years ago

Hydrogen gas, iodine vapor, hydrogen iodine are mixed in a flask and heated to 696°C. H2(g) + I2(g) ⇋ 2 HI(g) Kc = 52 at 696°C I

Anettt [7]

Answer: The equilibrium concentration of hydrogen gas is 0.0269 M

Explanation:

The chemical reaction follows the equation:

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

At t = 0 0.044M 0.044M 0.177M

At $t=t_{eq}$ (0.044-x)M (0.044-x)M (0.177+x)M

The expression for $K_c$ for the given reaction follows:

$K_c=\frac{[HI]^2}{[H_2]\times [I_2]}$

We are given:

$K_c=52$

Putting values in above equation, we get:

$52=\frac{(0.177+x)^2}{(0.044-x)^2}$

$x=0.0171M$

Hence, the equilibrium concentration of hydrogen gas is (0.044-x) M =(0.044-0.0171) M= 0.0269 M

7 0

3 years ago