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lys-0071 [83]
3 years ago
15

Classify the following as fluids or solids: warm butter, liquid nitrogen, paper, neon gas, ice.

Chemistry
1 answer:
gogolik [260]3 years ago
6 0
Warm butter- liquid
Liquid nitrogen- gas
paper- solid
neon gas- gas
ice- solid
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An organic compound has a distribution coefficient of 1.5
jasenka [17]

Answer:

Three times with 5 ml will yield more

Explanation:

Let x represent the amount yield

Kd = (x/15) / ((50-x) / 15) where Kd = 1.5

1.5 = (x/15) / ((50-x) / 15)

x / (50 - x) = 1.5

x = 75 - 1.5x

x + 1.5x  = 75

2.5x = 75

x = 75 / 2.5 = 30 mg

when extraction  three times

1st extraction

(x1/5) / ((50 - x1) / 15) = 1.5

3x1 / 50 - x1 = 1.5

3x1 = 75 - 1.5x1

3x1 + 1.5x1 = 75

4.5x1 = 75

x1 = 75 / 4.5 = 16.67 gm

second extraction

(x2/ 5) / (33.33 - x2 ) / 15) = 1.5

3x2 / ( 33.33 - x2) = 1.5

3x2 = 1.5(33.33 - x2)

3x2 = 49.995 - 1.5x2

3x2 + 1.5x2 = 49.995

4.5x2 = 49.995

x2 = 49.995 / 4.5

x2 = 11.11 mg

Third extraction

(x3/5) / ((22.22 - x3) / 15) = 1.5

3x3 = 1.5 ( 22.22 - x3)

3x3 + 1.5x3 = 33.33

4.5x3 = 33.33

x3 = 33.33 / 4.5 = 7.41 mg

total extraction = x1 + x2 + x3 =16.67 + 11.11 + 7.41 = 35.19 mg

the three times extraction using 5ml yields 5.19 mg more

3 0
3 years ago
Which of the elements must be available in water for the metamorphosis of tadpoles?
Grace [21]
The element iodine (I) is important for the fast and hastened metamorphosis of frog-tadpoles. As amphibians, the tadpoles can live in water and land but when they are born they are iodine-deficient. Tadpoles that do not receive ample amount of iodine become tadpoles until the end of their days. 

8 0
3 years ago
How does the law of conservation of matter apply to chemical equations?
Tasya [4]

Answer: option D.

The total number of atoms of each element on both sides of the

equation must be the same.

Explanation:

3 0
3 years ago
Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess
Tpy6a [65]

Answer:

The equilibrium constant in terms of concentration that is, K_c=3.6243\times 10^{3} .

Explanation:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The relation of K_c\& K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p= Equilibrium constant in terms of partial pressure.=98.1

K_c= Equilibrium constant in terms of concentration  =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

\Delta n_g = Difference between gaseous moles on product side and reactant side=n_{g,p}-n_{g.r}=1-2=-1

98.1=K_c(RT)^{-1}

98.1 =\frac{K_c}{RT}

K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}

The equilibrium constant in terms of concentration that is, K_c=3.6243\times 10^{3} .

5 0
3 years ago
Read 2 more answers
I need help answer with right answer
larisa86 [58]

Answer:

I think it would be D

Explanation:

6 0
3 years ago
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