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4 years ago

9

Which of the following describes the degree to which molecules of one substance are attracted to the molecules of another substa

nce?
1)surface tension
2)cohesion
3)adhesion
4)capillary action

2 answers:

DiKsa [7]4 years ago

8 0

The answer is adhesion.

Arada [10]4 years ago

5 0

The answer is adhesion

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Which of the following statements is false? Check all that apply. Gases are far more compressible than liquids The volume occupi

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Answer:

The false statement is: 3. Because liquid water and liquid carbon tetrachloride do not mix, neither do their vapors

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3 years ago

Differentiate between sandy soil, clayey soil and loamy soil on the basis of the growth of plants????

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The mixture of rock particle sand humus is called the soil.
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3 years ago

222 g of MTBE (CO(CH3)4) are added to gasoline, resulting in a total volume of 2 L of reformulated gas (RFG). Assume that the de

Eva8 [605]

Answer:

Explanation:

From the information given:

(a)

$Concentration \ in \ mg/L = \dfrac{Mass \ of \ MTBE \ in \ mg}{Total \ volume (in \ L)}$

$Concentration \ in \ mg/L = \dfrac{222 \times 10^3 \ mg}{22}$

$Concentration \ in \ mg/L = 111 \times 10^3 \ mg/L$

(b)

$number \ of \ mole s= \dfrac{mass}{molar \ mass } \\ \\ number \ of \ mole s=\dfrac{222 \ g}{88.15 \ g/mol} \\ \\ \mathbf{= 2.518 mol}$

(c)

$w/w \ percentage = \dfrac{mass \ of \ MTBE }{mass \ of \ solution (RFG)}\times 100\%$

$where; \\ \\ mass \ of \ (RFG) = 2L \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 2000 ml \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 1400 g$

∴

$w/w \ percentage = \dfrac{222 \ g}{1400 \ g}\times 100\% = \mathbf{15.8\%}$

(d)

$Volume of MTBE =\dfrac{mass \ of \ MTBE}{density \ of \ MTBE}$

$Volume \ of \ MTBE = 300 \ mL\\$

∴

$v/v\% = \dfrac{volume \ of \ MTBE}{volume \ of \ RFG} \\ \\ v/v\% =\dfrac{300 \ mL}{2000 \ mL}\times 100\% \\ \\ \mathbf{v/v\% = 15.00\%}$

(e)

$From \the \ given \ information; \\ \\ 2.5184 \ moles\ of \ MTBE contain \ 2.5184 \ mole of oxygen$

∴

$mass of oxygen MTBE = 2.5284 mol \times 16\ g/mol \\ \\ mass of oxygen MTBE = 40.3 9 \ g\\ \\ mass\ of \ RFG = 1400 g$

∴

$\% w/w = \dfrac{mass \ of \ oxygen}{mass \ of RFG }=\dfrac{40.22 \ g}{1400 \ g} \times 100\%$

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3 0

3 years ago

Read 2 more answers

How many particles are in 7.07x10^-6 moles of a substance?

Butoxors [25]

Answer:

4.25*10^18

Explanation:

1 mole =6.023*10^23 particles

so 7.07*10^-6 mole=6.023*10^23*7.07*10^-6 particles=4.25*10^18 particles

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3 years ago

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ivolga24 [154]

This type of reaction is called: Synthesis

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