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Harlamova29_29 [7]
3 years ago
10

How is oxygen and nitrogen alike​

Chemistry
2 answers:
Viefleur [7K]3 years ago
5 0

Answer:they both if we mix them they explosive

Explanation:because its two dangerous

11111nata11111 [884]3 years ago
3 0

Answer:

They are both colorless, odorless, and tasteless. They have the same number of valence electrons too. And unbalanced electrons in their valence shell.

Explanation:

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HELP WITH CHEMISTRY PLEASE!
maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

  • In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
  • Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

<u><em>In our problem:</em></u>

P1 = ??? <em>(is needed to be calculated) </em>and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.


2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

<u><em>In our problem:</em></u>

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? <em>(is needed to be calculated) </em>

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.


3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

<u><em>In our problem:</em></u>

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? <em>(is needed to be calculated) </em> and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.


4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; <em><u>101.325 kPa = 1.0 atm</u></em>, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.


5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

<u><em>In our problem:</em></u>

P1 = 2200.0 mmHg and T1 = ??? <em>(is needed to be calculated) </em>.

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

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Due to lava which heats water where water vapour is comes
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  <span>C. polar bonds and asymmetrical structure 
If the molecule contains polar bonds but it has a symmetrical structure, the polar bonds will cancel each other out so the overall molecule will be non-polar. 
On the other hand, if the molecule contains polar bonds but has an asymmetrical structure, then the polar bonds won't cancel each other out, so the overall molecule ends up being polar. 
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Which of the following would NOT diffuse through the plasma membrane by means of simple diffusion?1 oxygen2 glucose3 a steroid h
Mumz [18]

Answer:

Option 2= Glucose

Explanation:

Cell membrane is made up of two phospholipid layers and each contain phosphate head and fatty acid or lipid tails. the head is present between the outer and inner boundaries and tail is present in between. The small non- polar molecules can pass the membrane through simple diffusion. This lipid tail restrict the passage of polar molecules including water soluble substances like glucose. However, transmembranes are present that allow the molecules to inter that are blocked by the tails.

Facilitated diffusion:

it is a type of diffusion in which caries protein without using the cellular energy shuttle the molecules to the cell membrane. Glucose is bind on the carrier protein ,change the shape and transport it from one to another side of membrane. In order to absorb the glucose red blood cells use this kind of diffusion.

Primary active transport:

The cells that are present along small intestine use this type of transport to pump the glucose inside the cell. The primary active transport require energy to transport the glucose inside.

Secondary active transport:

It is another method of transport of glucose into the cell. This method can not use ATP but it is based on concentration gradient of the sodium that provide electro chemical energy for the glucose transport.

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