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Molodets [167]
4 years ago
12

Sweat is a mixture of water, dissolved solids, and chemicals. How does the water become separated from the sweat leaving a perso

n's skin salty? A. It separates naturally. B. It separates due to density. C. It separates by evaporation. D. It separates due to magnetism.
Chemistry
2 answers:
zalisa [80]4 years ago
5 0

The answer is C. because evaporation

Nataly_w [17]4 years ago
4 0
C) it separates due to evaporation
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What task can a cell NOT perform on its own?
Pachacha [2.7K]
Thought cannot be performed or produced by a cell as it is devoid of a capable nerve system and hence it does not get any information and thus cannot produce any thought.
5 0
3 years ago
NO LINKS RIGHT ANSWERS ONLY!!!!
Mumz [18]
I think the answer is D
6 0
3 years ago
Read 2 more answers
How to do limiting reagents for chemistry?
gavmur [86]
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
4 0
3 years ago
Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o
stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0
3 years ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
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