Separating Mixtures Checkpoint ( giving brainly )

Which of the following liquids would have the highest viscosity at room temperature?

Liono4ka [1.6K]

Answer:

5)HOCH2CH2OH

Explanation:

This is also known as ethylene glycol. An increase in hydrogen bonds of a compound means an increase in the viscosity. Hydrogen bonds occur as a result of bonding with electronegative elements such as Oxygen, Nitrogen etc.

The compounds with the highest amount of Hydrogen bond represents the one with the highest viscosity which is B) HOCH2CH2OH

3 0

2 years ago

1. Which statements represent properties of intermolecular forces? Select all that apply.

kumpel [21]

Answer:

C

Explanation:

Only this choice is applicable and correct.

The inter-molecular forces of attraction between the molecules must have been broken (overcome) before the molecules can gain an increase in the kinetic energies between them.

8 0

2 years ago

Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. The balanced chemical equation

Rufina [12.5K]

The mass formed in a chemical reaction is given by stoichiometry law in balanced chemical equation. The mass of silver produced in the reaction is 107. 9 g.

<h3>What is the stoichiometric law?</h3>

The stoichiometric law states that in a balanced chemical equation, the stoichiometric coefficient describes the moles of each compound in a chemical reaction.

The balanced chemical equation for the reaction is:

$\rm Cu\;+\;2\;AgNO_3\;\rightarrow\;Cu(NO_3)_2\;+\;2\;Ag$

From the balanced chemical equation, 1 moles of copper results in 2 moles of silver.

Moles in terms of mass can be given as:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

Thus, moles of 31. 75 grams copper is:

$\rm Cu=\dfrac{ 31. 75 }{63.5}\;mol\\ Cu=0.5 mol$

The moles of Ag produced can be given from stoichiometric law as:

$\rm 1\;mol\;Cu=2\;mol\;Ag\\0.5 \;mol\;Cu=0.5\;\times\;2\;mol\;Ag\\0.5\;mol\;Cu=1\;mol\;Ag$

The moles of Ag produced is 1 mol.

The mass of a mole of element is equivalent to the molar mass. Thus, the mass of Ag produced in the reaction is 107. 9 g. Hence, option B is correct.

Learn more about stoichiometric law, here:

brainly.com/question/6907332

7 0

1 year ago

4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.

slava [35]

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ $\frac{1mol}{102g}$ = 0,0980 moles

And 10,0g of HCl are:

10,0 gₓ $\frac{1mol}{36,5g}$ = 0,274 moles

For a total reaction of 0,274 moles of HCl you need:

0,274× $\frac{1molesAl_{2}O_3}{6 mole HCl}$ = 0,0457 moles of Al₂O₃

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

0,274 moles HCl × $\frac{2 moles AlCl_{3}}{6 moles HCl}$ × 133 $\frac{g}{mol}$ = 12,1 g of AlCl₃

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles = 0,0523 moles

And its mass is:

0,0523 molesₓ $\frac{102g}{1mol}$ = 5,33 g of Al₂O₃

I hope it helps!

7 0

2 years ago

Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.30 L and the temperature is increased to 4

lakkis [162]

Answer:

P₂ = 1.12 atm

Explanation:

To find the new pressure, you need to use the Combined Gas Law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the new pressure, volume, and temperature. Before plugging the values into the equation, you need to

(1) convert the pressure from mmHg to atm (760 mmHg = 1 atm)

(2) convert the temperatures from Celsius to Kelvin (°C + 273)

The final answer should have 3 sig figs like the given values.

P₁ = 365 mmHg / 760 = 0.480 atm P₂ = ? atm

V₁ = 5.00 L V₂ = 2.30 L

T₁ = 20°C + 273 = 293 K T₂ = 40°C + 273 = 313 K

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ <----- Combined Gas Law

$\frac{(0.480 atm)(5.00 L)}{293 K}=\frac{P_2(2.30 L)}{313 K}$ <----- Insert values