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VLD [36.1K]
3 years ago
14

Identify the mixture of gasoline and motor oil and suggest a technique for separating their components. 1. homogeneous; decantin

g 2. heterogeneous; decanting 3. heterogeneous; distillation 4. homogeneous; distillation 5. homogeneous; dissolving followed by filtration and distillation
Chemistry
2 answers:
uranmaximum [27]3 years ago
7 0
Greese mixed liquid petroleum use in gasoline
alexgriva [62]3 years ago
4 0

Answer:

4. homogeneous; distillation

Explanation:

Gasoline and motor oil are chemically similar. They are both mixtures of non polar hydrocarbons containing carbon and hydrogen atoms. However, motor oil is much more viscous Motor Oil. Hence we can say that the mixture of gasoline and motor oil are homogeneous and they can be separated by distillation.

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Help How many atoms of hydrogen(gas) form if 10 grams of aluminum react? 2Al + 6HCl = 2AlCl3 + 3H2
arlik [135]

Answer: 3.35x10²³atoms H2

Explanation: solution attached:

Convert mass of Al to moles

Do the mole to mole ratio between Al and H2

Convert moles of H2 to atoms using Avogadro's number.

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Hello Friends Can Yall Plz Help Me For Brainliest Don’t Comments No B.S Yu Will Get Cursed out show ya work
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2 years ago
Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)
marin [14]

Answer:

The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

Explanation:

Suppose there are 100 moles in solution:

Moles of sulfuric acid = 6% of 100 moles = 6 moles

Mass of 6 moles of sulfuric acid = 6 mol × 98 g/mol=588 g

Moles of water = 100%- 6% = 94%= 94 moles

Mass of water = 94 mol × 18 g/mol = 1692 g

Specific gravity of the solution ,S.G= 1.07

Density of solution = D

S.G=\frac{D}{d_w}

d_w = density of water = 1 g/mL

D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL

Mass of the solution = 588 g + 1692 g = 2280 g

Volume of the solution = V

Volume = \frac{Mass}{Density}

=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L

1 mL = 0.001 L

Molarity = \frac{n}{V(L)}

n = number of moles of compound

V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L

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2 years ago
El ácido sulfúrico H2SO4 es uno de los compuestos que se utiliza para la producción de fertilizantes como el nitrosulfato amónic
Serjik [45]

Explanation:

12 hours ago

El ácido sulfúrico H2SO4 es uno de los compuestos que se utiliza para la producción de fertilizantes como el nitrosulfato amónico. Si disponemos de 8 mL de H2SO4 al 37 %P/P (d=1,26 g /mL), los cuales se disolvieron hasta alcanzar un volumen de solución de 400 mL, con una densidad de 1,08 g/mL. (La densidad del soluto es corresponde a 1,83 g/cm³)

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3 years ago
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