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wolverine [178]
2 years ago
5

I will give 25 points and make them brainlest if they solve this.

Chemistry
2 answers:
sdas [7]2 years ago
7 0

Answer:

B

Explanation:

serious [3.7K]2 years ago
6 0

Answer:

B - Carbon atoms found in current living things are the same carbon atoms that have always been a part of living things, they are just recycled.

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Who wants to play truth or dare
natima [27]

Iam willing to play truth and dare

Explanation:

because I love to play truth and dare

6 0
2 years ago
Read 2 more answers
Use Boyle’s law to complete the following:
mixas84 [53]

Answer:

The answer to your question is: 0.25 l

Explanation:

Data

P1 = 1 atm

V1 = 0.5 l

P2 =2 atm

V2 = ?

T = constant

Formula

          V1P1 = V2P2

Clear V2 from the formula

            V2 = V1P1/P2

Substitution

            V2 = (0.5)(1)/2  substitution

                  = 0.25 l       result

3 0
3 years ago
PLEASEE ANSWERRR FASTTT
IceJOKER [234]

Answer:

last one

Explanation:

The elements classified as metalloids are boron, silicon, germanium, arsenic, antimony, tellurium, and polonium.

8 0
3 years ago
Read 2 more answers
If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of 200 K and then I raise the pressu
mixas84 [53]

Answer:

The answer to your question is V2 = 29.6 l

Explanation:

Data

Pressure 1 = P1 = 12 atm

Volume 1 = V1 = 23 l

Temperature 1 = T1 = 200 °K

Pressure 2 = 14 atm

Volume 2 = V2 = =

Temperature 2 = T2 = 300°K

Process

1.- To solve this problem use the Combine gas law.

             P1V1/T1 = P2V2/T2

-Solve for V2

             V2 = P1V1T2 / T1P2

2.- Substitution

             V2 = (12)(23)(300) / (200)(14)

3.- Simplification

             V2 = 82800 / 2800

4.- Result

            V2 = 29.6 l

5 0
3 years ago
The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e
hoa [83]

Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

                         x = (3.268 x 12) / 44

                        x = 0.891 g of Carbon

                       12 g of carbon -----------  1 mol

                       0.891 g of C     ----------   x

                       x = (0.891 x 1) / 12

                       x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

                      1 g of hydrogen ------------  1 mol of H

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

                      x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

                           = 0.299 g of O₂

Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

                             x = 0.019 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon         0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

4 0
3 years ago
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