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3 years ago

13

Explain the relationship between chemical bonds and energy in the body.

1 answer:

uysha [10]3 years ago

6 0

It is a very simple explanation. When the chemical bonds are broken inside our body, the energy that gets released in our body is used to perform several important functions. This is the main relation between chemical bonds and energy in the body. I hope that this is the answer that has come to your great help.

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You are working in the lab, how many grams of solute you need to make 200.0 g of 3.4% of solution?

Goryan [66]

Answer:

6.8g of solute are needed

Explanation:

Percent by mass, (w/w%) is defined as the mass of solute in 100g of solution. A solution that is 3.4% contains 3.4g of solute in 100g of solution. That means to make 200g of solution are required:

200g solution * (3.4g solute / 100g solution) = 6.8g of solute are needed

6 0

3 years ago

What are the particles of an atom called that are in the outermost energy level?

densk [106]

The atoms on the outermost level are called electrons.

8 0

2 years ago

2Li+2H2O—>2LiOH+H2

taurus [48]

Explanation:

2Li+2H2O—>2LiOH+H2

Calculate the mass of reacted lithium when H2 is 6.02 * 10 ^ 23 molecules.

I really need the answer with all the calculation please.

3 0

2 years ago

Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after

Alenkinab [10]

<u>Answer:</u>

<u>For a:</u> The number of moles of air present in the RV is 0.047 moles

<u>For b:</u> The number of molecules of gas is $2.83\times 10^{22}$

<u>Explanation:</u>

<u>For a:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air = $37^oC=[37+273]K=310K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of air = ?

Putting values in above equation, we get:

$1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol$

Hence, the number of moles of air present in the RV is 0.047 moles

<u>For b:</u>

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.047 moles of air will contain $(0.047\times 6.022\times 10^{23})=2.83\times 10^{22}$ number of gas molecules.

Hence, the number of molecules of gas is $2.83\times 10^{22}$

7 0

2 years ago

Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.

Oduvanchick [21]

Answer:

103.9 g

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:

54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:

1.23 mol C₃H₈ * $\frac{3molCO_2}{1molC_3H_8}$ = 3.69 mol CO₂

We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

3.69 mol CO₂ * 44 g/mol = 162.36 g

And <u>apply the given yield</u>:

162.36 g * 64.0/100 = 103.9 g

7 0

2 years ago