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nika2105 [10]
2 years ago
10

Drag the labels onto the flowchart to indicate how food molecules reach the body’s cells and fuel cellular respiration. start wi

th the ingestion of food on the left.
Chemistry
1 answer:
kogti [31]2 years ago
3 0
Start with the ingestion of food on the left. Eating food provides fuel & building blocks for your body, after food is broke down in the digestive system, it is transported to cells via the circulatory system, fuel molecules are broken down further in glycolysis& the citric acid cycle, ATP is produced with <span>the help of the electron transport chain.</span>
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Write an equation that shows the formation of a strontium ion from a neutral strontium atom
Pepsi [2]

Answer:

See explanation

Explanation:

Sr(s) + 2HCl(aq) -----> SrCl2(aq) + H2(g)

Ionically;

Sr(s) + 2Cl^-(aq) ----> SrCl2(aq)

If we look at the reaction above, strontium atom was dissolved in hydrochloric acid. The strontium atom is now oxidized by the acid to give Sr^2+ ion according to the equation shown above.

3 0
2 years ago
A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w
MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}

Mass of the water =1 g/mL\times 165 mL=165 g

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%

The weight/weight % or percent by mass of the solute is 5.41 %.


8 0
2 years ago
A mixture of carbon dioxide and an unknown gas was allowed to effuse from a container. The carbon dioxide took 1.25 times as lon
Aleks04 [339]

Answer:

CO

Explanation:

From Graham's law, time taken to diffuse is directly proportional to the molecular mass of the gases. For two different gases.

t1/t2=√m1/m2

Since gas 1 diffuse 1.25 times as slowly as gas 2 and gas 1 is CO2 with m as 44g

1.25/1=√44/m2

Therefore m2=28g CO

7 0
2 years ago
The concentration of ozone in a sample of air that has a partial pressure of O3 of 0.33 torr and a total pressure of air of 695
Goryan [66]

Answer:

0.047 %

Explanation:

Step 1: Given data

  • Partial pressure of ozone (pO₃): 0.33 torr
  • Total pressure of air (P): 695 torr

Step 2: Calculate the %v/v of ozone in the air

Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.

<em>%v/v = pO₃/P × 100%</em>

%v/v = 0.33 torr/695 torr × 100%

%v/v = 0.047 %

8 0
2 years ago
What is the benefit of having a limiting reagent when performing a lab experiment
Shalnov [3]

Answer : The role of limiting reagent or reactant is important in a chemical reaction because it can help the chemist to predict that complete amount of reactant is consumed, as it is limiting the reaction, only required moles of products can get formed instead of the theoretical yield where the perfect amount is used.


In short, Limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is found to be complete.

3 0
2 years ago
Read 2 more answers
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