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svetoff [14.1K]
3 years ago
9

Why is oil on able to dissolve well in water?

Chemistry
2 answers:
Igoryamba3 years ago
6 0
Hello there.

<span>Why is oil on able to dissolve well in water?

</span><span>B. Oil molecules or too large to fit between the closely spaced water molecules in the liquid state 
</span>
chubhunter [2.5K]3 years ago
3 0
Its the answer b, because oil has a great density, therefore, when you mix them together the oil will be below and the above of it. 

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How many grams in 11.9 moles of sulfur? Avogadro’s number: 6.02x1023 atoms = 1 mole Molar mass of sulfur: 32.06 g sulfur = 1 mol
sergiy2304 [10]

Answer:

Mass = 0.37 g

Explanation:

Given data:

Number of moles of sulfur = 11.9 mol

Mass of sulfur in 11.9 mol = ?

Molar mass of sulfur = 32.06 g

Solution:

Number of moles = mass/molar mass

by putting values,

11.9 mol = mass/ 32.06 g/mol

Mass = 11.9 mol × 32.06 g/mol

Mass = 0.37 g

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When one atom of carbon reacts with two atoms of oxygen, they combine to form one molecule of carbon dioxide. How many atoms of
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The equation is: C+O2=>CO2

Since we got 10 molecules of CO2 new balanced equation would be 10C+10O2=>10CO2

from this equation we can see that we have 10 molecules of oxygen, however ,we need to find atoms. There are 2 atoms in the oxygen molecule so we need to multiply 10 by 2 which gives us 20 atoms.

The answer: there are 20 atoms of oxygen

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For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The
babymother [125]

Answer : The order of reaction with respect to A is, second order reaction.

The order of reaction with respect to B is, zero order reaction.

The order of reaction with respect to C is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Thus,

The order of reaction with respect to A is, second order reaction.

The order of reaction with respect to B is, zero order reaction.

The order of reaction with respect to C is, first order reaction.

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3 years ago
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