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vovikov84 [41]
2 years ago
12

Which energy transformation produces electricity and is least likely to negatively impact the environment?

Chemistry
1 answer:
Anastasy [175]2 years ago
8 0

<em>Hello!</em>

<h3><em /></h3><h3><u><em>The answer to this question is</em></u>: D) Generating electricity from a solar panel system.</h3><h3 />

<em><u>Thank you for asking this question!</u></em>

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<em><u /></em>

<em><u>It would be greatly appreciated if you thanked me and marked me as the brainliest!</u></em>

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<em><u>Have an amazing day.</u></em>

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Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di
katrin [286]

<u>Answer:</u> The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

<u>Explanation:</u>

  • <u>Calculating the molarity of solution:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M

  • <u>Calculating the molarity of solution:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g

To calculate the molality of solution, we use the equation:

\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

m_{solute} = Given mass of solute (camphor) = 70 g

M_{solute} = Molar mass of solute (camphor) = 152.2  g/mol

W_{solvent} = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m

  • <u>Calculating the mole fraction of camphor:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For camphor:</u>

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol

<u>For ethanol:</u>

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

\chi_{(camphor)}=\frac{0.459}{10.272}=0.045\

  • <u>Calculating the mass percent of camphor:</u>

To calculate the mass percentage of camphor in solution, we use the equation:

\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0
3 years ago
100.0 mL of Ca(OH)2 solution is titrated with 5.00 x 10–2 M HBr. It requires 36.5 mL of the acid solution for neutralization. Wh
miskamm [114]

Answer:

The number of moles HBr = 0.001825

The concentration of Ca(OH)2 = 0.009125 M

Explanation:

Step 1: Data given

Volume of the Ca(OH)2 = 100.0 mL = 0.100 L

Molarity of HBr = 5.00 * 10^-2 M

Volume of HBR = 36.5 mL = 0.0365 L

Step 2: The balanced equation

Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Step 3: Calculate molarity of Ca(OH) 2

b*Va* Ca = a * Vb*Cb

⇒with b = the coefficient of HBr = 2

⇒with Va = the volume of Ca(OH)2 = 0.100 L

⇒with ca = the concentration of Ca(OH)2 = TO BE DETERMINED

⇒with a = the coefficient of Ca(OH)2 = 1

⇒with Vb = the volume of HBr = 0.0365 L

⇒with Cb = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Ca = (0.0365*0.05) / 0.200

Ca = 0.009125 M

Step 4: Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

3 0
3 years ago
Sometimes, during the DNA replication process, mistakes are made. These are called
wolverine [178]

Answer: IT'S MUTATION I TOOK THE TEST ITS RIGHT HOPE THIS HELPED :)

8 0
3 years ago
Read 2 more answers
Please refer to image please
GarryVolchara [31]

Answer:

76.9L

Explanation:

Based on the graph, whenever the temperature increases by 100K, the volume increases by 10L, so do 769/10

7 0
2 years ago
3.80 moles of oxygen are used up in the reaction. How many moles of water are produced?5.How many grams of oxygen does it take t
r-ruslan [8.4K]

Answer:

2 KClO3 (s) = 2 KCl (s) + 3 O2 (g)

2.5 g x g

Explanation:

x g O2 = 2.5 g KClO3 x (1 mol KClO3) x (3 mol O2) x (32 g O2) = 0.98 g O2

(122.5 g KClO3) (2 mol KClO3) (1 mol O2)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

2.5 g x g

x g KCl = 2.5 g KClO3 x (1 mol KClO3) x (2 mol KClO3) x (74.5 g KCl) = 1.52 g KCl

(122.5 g KClO3) (2 mol KClO3) (1 mol KCl)

2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

x mol 10 mol

x mol KClO3 = 10 mol O2 x (2 mol KClO3) = 6.7 mol KClO3

(3 mol O2)

7 0
3 years ago
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