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2 months ago

What is the mass in grams of 16.3 mol of the element nickel, Ni?

Chemistry

1 answer:

Alekssandra [3K]2 months ago

4 0

The weight of 16.3 moles of nickel amounts to 956.647 g.

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A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

KiRa [2933]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

Assuming mixing doesn't alter the total volume.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

4 0

2 months ago

Every single-celled organism is able to survive because it carries out *

castortr0y [3046]

Every unicellular organism prospers by executing metabolic activities.

Metabolic activities encompass the set of chemical reactions essential for sustaining life.

Explanation:

Different metabolic pathways maintain an organism's viability. Various metabolic activities occur in all living organisms.

These include processes like cellular respiration, reproduction, excretion, and digestion. Each living cell engages in these activities to survive.

Organisms acquire the energy necessary for these activities through food consumption.

Read more on -

6 0

2 months ago

At the boiling point, the density of the liquid is 809 g/l and that of the gas is 4.566 g/l. how many liters of liquid nitrogen

KiRa [2933]

Result: 1.68 L of liquid nitrogen is generated during the gas liquefaction process.

Clarification:

This process involves transforming gaseous nitrogen into its liquid form.

The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.

Step 1: Calculate the mass of nitrogen gas

Let’s determine the mass of nitrogen gas associated with 297 L.

The density formula is:

$Density = \frac{Mass}{Volume}$

With a density of nitrogen gas at 4.566 g/L and a volume of 297 L, we can compute the mass of nitrogen gas as follows:

Using these values yields:

$Mass = Density \times Volume$

$Mass = \frac{4.566g}{L} \times 297L$

$Mass = 1356g$

The mass of nitrogen gas calculates to be 1356 g.

Step 2: Derive the volume of liquid nitrogen from the mass obtained

The mass for liquid nitrogen remains the same.

With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.

$Volume = \frac{Mass}{Density}$

$Volume = \frac{1356g}{809g/L}$

Therefore, the volume of liquid nitrogen is 1.68 L.

6 0

2 months ago

Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversi

Tems11 [2777]

Answer:

15.71g

Explanation:

The combustion equation that applies to hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

In the case of octane, C8H18:

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (this is the balanced equation)

From this balanced reaction,

2 x 22.4 L of octane generates 16 [ 12 + (16 x 2)] of carbon dioxide

That means,

44.8 L of octane generates 704g of carbon dioxide

Thus, for 1L of octane, it produces 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Consequently, 15.71g of carbon dioxide is produced from the complete combustion of 1 L of octane.

7 0

2 months ago