Name two useful materials obtained from fractional distillation of petroleum

Calculate the mass percent of HOCH 2CH 2OH in a solution made by dissolving 3.2 g of HOCH 2CH 2OH in 43.5g of water.

grin007 [14]

Answer:

$\%m/m=6.85\%$

Explanation:

Hello,

In this case, we are asked to compute the by mass percent for the given 3.2 g of ethylene glycol in 43.5 g of water. In such a way, since the by mass percent is computed as follows:

$\%m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%$

Whereas the solute is the ethylene glycol and the solvent the water, therefore we obtain:

$\%m/m=\frac{3.2g}{3.2g+43.5g} *100\%\\\\\%m/m=6.85\%$

Best regards.

5 0

3 years ago

A LA TEMPERATURA DE 35°C, UNA MUESTRA DE DIOXIDO DE CARBONO OCUPA UN VOLUMEN DE 350 ML. ¿Qué CAMBIO DE VOLUMEN SE PRODUCIRA SI L

Kisachek [45]

Answer:

New volume = 150 mL

Explanation:

Initial temperature, T₁ = 35°C

Initial volume, V₁ = 350 mL

We need to find the change in volume when the temperature drops to 15°C.

The relation between the temperature and the volume is given by Charle's law. Let new volume is V₂. It can be given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{350\times 15}{35}\\\\V_2=150\ mL$

So, the new volume is 150 mL.

8 0

3 years ago

C(s) + S8(s) → CS2(l)

Vanyuwa [196]

The balanced chemical reaction is written as:

4C(s) + S8(s) → 4CS2(l)

We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.

7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8

The limiting reactant would be S8. We use this amount to calculate.

0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2

4 0

3 years ago

This question has multiple parts. Work all the parts to get the most points. An aqueous antifreeze solution is 39.0% ethylene gl

nlexa [21]

Answer:

Molality = 10.300 m

Molarity = 6.5970 M

mole fraction = 0.156549

Explanation:

39.0 % = ethylene glycol

61.0 % = water

imagine the total mass = 100g

39.0% ethylene glycol = 39g

61.0 % water = 61g

1) Molality = number of moles / mass of solvent (kg)

Molar mass of ethylene glycol = 62.07g/mole

mole = 39g / 62.07g/mole = 0.6283 moles

Molality = moles / mass of solvent = 0.6283 moles / 0.061kg = 10.300 m

2) Molarity = number of moles / volume of solution

Since we know the density of the solution = 1.05g /ml

⇒ volume = 100g / 1.05g /mL = 95.24 mL = 0.09524 L

Molarity = 0.6283 moles / 0.09524 L = 6.5970 M

3) Mole fraction

moles water = 61g / 18.02g/mole = 3.38513 moles

Total number of moles = moles of ethylene glycol + moles of water = 0.6283 + 3.38513 = 4.0134276 moles

Mole fraction = 0.6283/ 4.0134276 = 0.156549

3 0

3 years ago

What is the net ionic equation of the reaction of mgso4 with sr(no3)2?

Bad White [126]

Balanced chemical reaction:

MgSO₄(aq) + Sr(NO₃)₂(aq) → Mg(NO₃)₂(aq) + SrSO₄(s).

Ionic reaction:

Mg²⁺(aq) + SO₄²⁻(aq) + Sr²⁺(aq) + 2NO₃⁻(aq) → Mg²⁺(aq) + 2NO₃⁻(aq) + SrSO₄(s).

Net ionic reaction:

Sr²⁺(aq) + SO₄²⁻(aq) → SrSO₄(s).

Magnesium sulfate (MgSO₄), strontium nitrate (Sr(NO₃)₂ and magnesium nitrate (Mg(NO₃)₂) are soluble in water. Strontium sulfate (SrSO₄) is not soluble in water.

This chemical reaction is double displacement reaction - cations and anions of the two reactants switch places and form two new compounds.

6 0

3 years ago

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