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Anettt [7]
3 years ago
10

Can you use active transport in a sentence

Chemistry
2 answers:
anzhelika [568]3 years ago
8 0
Active transport is the movement of a substance agasinst its concentration energy. Hope this helps. 
otez555 [7]3 years ago
6 0

Active transport is the moving of molecules across the membrane of the cell against the concentration gradient with the use of ATP.

Low to high concentration. Concentration gradient is the diffusion (movement of molecules from regions of low concentration) from high to low with the gradient. Active transport is from low to high, against the gradient.

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The following data is given to find the formula of a Hydrate:
umka21 [38]

The masses can be found by substractions:

  • Mass of CaSO₄.H2O (hydrate):

16.05 g - 13.56 g = 2.49 g

  • Mass of CaSO₄ anhydrate:

15.07 g - 13.56 g = 1.51 g

  • The mass of water is equal to the difference between the mass of the hydrate and the mass of the anhydrate:

2.49 g - 1.51 g = 0.98 g

  • The percent of water is found by the formula:

massWater ÷ massHydrate * 100%

0.98 g ÷ 2.49 g * 100% = 39.36%

  • The mole of water is calculated using water's molecular weight (18g/mol):

0.98 g ÷ 18 g/mol = 0.054 mol water

  • A similar procedure is made for the mole of salt (CaSO₄ = 136.14 g/mol)

1.51 g ÷ 136.14 g/mol = 0.011 mol CaSO₄

  • The ratio of mole of water to mole of anhydrate is:

0.054 mol water / 0.011 mol CaSO₄ = 0.49

In other words the molecular formula for the hydrate salt is CaSO₄·0.5H₂O

3 0
3 years ago
What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o
Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g)  → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0
3 years ago
The unique odors and flavors of many fruits are primarily due to small quantities of a certain class of organic compounds. The e
SpyIntel [72]
Based on the diagram shown, a numerical setup for calculating the gram-formula mass for reactant 1 would be :
6(1) + 2(12) + 16

Hope this helps
5 0
3 years ago
The burning of methane gas, given below, is a redox reaction. which part of the reaction illustrates oxidation?
Effectus [21]
The burning of methane gas, given below, is a redox reaction. which part of the reaction illustrates oxidation, Ch4+O2---CO2+H2O<span>CH4---CO2</span>
8 0
3 years ago
0.500 L of a gas is collected at 2911 MM and 0°C. What will the volume be at STP?
ioda

Answer:

V₂ =  1.92 L

Explanation:

Given data:

Initial volume = 0.500 L

Initial pressure =2911 mmHg (2911/760 = 3.83 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

by putting values,

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm

V₂ = 522.795 atm .L. K / 273 K.atm

V₂ =  1.92 L

4 0
3 years ago
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