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3 years ago

7th grade science , please don’t get the answer from the internet :(

Chemistry

1 answer:

Liula [17]3 years ago

7 0

Answer:

See the answer below

Explanation:

1. Organisms produce energy for cells by chemically breaking down and unlocking the energy locked-up within food materials in a process known as cellular respiration. The unlocked energy is then utilized for the cell's metabolic activities.

2. Cellular respiration can be aerobic or anaerobic.

Aerobic respiration involves the breakdown of carbohydrates in the presence of oxygen to yield energy in the form of ATP while carbon dioxide and water are produced as by-products.

$C_6H_1_2O_6 + 6O_2 --> 6CO_2 + 6H_2O + ATP$

Anaerobic respiration involves the breakdown of carbohydrates in the absence of oxygen to produce ATP and lactic acid as a by-product. The lactic acid is later oxidized to carbon dioxide and water to prevent it from building up.

$C_6H_1_2O_6 -> 2C_3H_6O_3 + ATP$

3. Photosynthesis and cellular respiration are both considered metabolic processes that take place in living organisms. However, photosynthesis is peculiar only to green plants and some algae while respiration is common to all living organisms. While photosynthesis is anabolic, that is, it involves the building up of materials; respiration is said to be catabolic because it involves the breaking down of materials.

During photosynthesis, inorganic products are utilized to produce carbohydrates for plants with oxygen gas released as a by-product according to the following equation:

$6CO_2 + 6H_2O --> C_6H_1_2O_6 + 6O_2$

During respiration, the food taken by living organisms is broken down to unlock the energy in it for metabolic activities according to the following equation:

$C_6H_1_2O_6 + 6O_2 --> 6CO_2 + 6H_2O + ATP$

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Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t

iragen [17]

Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0

C +S +S

E S S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

Solubility in 0.0120 M CoBr₂ (S')

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0.0240

C +S' +S'

E S' 0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

8 0

3 years ago

g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron

Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O Oxidation

BrO₃⁻ → Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ Reduction

In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0

3 years ago

Write the two balanced equations, one for the incomplete combustion of methane (to form carbon monoxide and water vapor), and an

kompoz [17]

Answer:

$2CH_{4(g)} +3O_{2(g)} -->2CO(g)+4H_{2}O_{(g)}\\CH_{4(g)} +2O_{2(g)} -->CO_{2(g)} +2H_{2}O_{(g)} \\$

Explanation:

$2CH_{4(g)} +3O_{2(g)} -->2CO(g)+4H_{2}O_{(g)}\\CH_{4(g)} +2O_{2(g)} -->CO_{2(g)} +2H_{2}O_{(g)} \\$

The incomplete combustion of alkanes, and other fuels actually, happens when there is a limited supply of oxygen. Instead of the fuel burning completely to produce carbon dioxide, it produces carbon monoxide instead.

This gas is harmful to jumans because it combines with haemoglobin in lood and takes up space that belongs to oxygen which can lead to suffocation or even death

6 0

3 years ago

True or false natural elements don't reflect light.

Burka [1]

I think the statement is false. Natural elements do reflect light. Almost everything in this world reflect light. Reflection is when light bounces off an object. Hope this answers the question. Have a nice day.

3 0

3 years ago

Why does the chemical reaction seen here obey the law of conservation of matter?

ss7ja [257]

Why does the chemical reaction seen here obey the law of conservation of matter?

Answer:

Because there are the same number of atoms of each element shown on both sides

Explanation

As in chemical reactions, atoms bonds are break and new bonds are formed. As new substance are formed but overall they have same elements, no new elements come from outside or go to outside. In other words , rearrangement of atoms take place but number of atoms remained same.

NaOH + HCl -----> NaCl + H2O

As in above reaction there are the same number of atoms of each element shown on both sides .

7 0

3 years ago

Read 2 more answers