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Flauer [41]
2 years ago
9

Air is about 78% nitrogen gas (N2) by mass. The molar mass of nitrogen gas is 28. 0 g/mol. A 100. 0-g sample of air contains how

many moles of nitrogen? 2. 8 3. 6 28 78.
Chemistry
1 answer:
andrezito [222]2 years ago
4 0

Answer:

fsadfd

Explanation:

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What is the mass sample of 0.0500 moles of zinc chloride ?
vlabodo [156]

Answer:

6.82g

0.59moles

Explanation:

1. What is the mass sample of 0.0500 moles of zinc chloride ?

Given parameters:

Number of moles ZnCl₂ = 0.05moles

Unknown:

Mass of the sample  =  ?

Solution:

To find the mass of a substance using the number of moles, it would be pertinent to understand what mole is.

A mole is a substance that contains the avogadro's number of particles.

It relates to the mass using the expression below;

                Mass of a substance  = number of moles x molar mass

Molar mass of  ZnCl₂;

        Atomic mass of Zn  = 65.4g/mol

                                   Cl = 35.5g/mol

Molar mass = 65.4 + 2(35.5)  = 136.4g/mole

Mass of a substance  = 0.05 x 136.4  = 6.82g

2. How many moles of potassium sulfide are in a 65.50g sample?

Given parameters:

Mass of K₂S  = 65.5g

Unknown:

Number of moles  = ?

Solution:

The number of moles of any substance is related to mass using the expression below;

              Number of moles  = \frac{mass}{molar mass}

Molar mass of K₂S  = 2(39) + 32  = 110g/mol

              Number of moles  = \frac{65.5}{110}   = 0.59moles

8 0
3 years ago
The kinetic molecular theory states that all particles of an ideal gas are
Marizza181 [45]

THE KINETIC MOLECULAR THEORY STATES THAT ALL PARTICLES OF AN IDEAL GAS ARE IN CONSTANT MOTION AND EXHIBITS PERFECT ELASTIC COLLISIONS.

Explanation:

An ideal gas is an imaginary gas whose behavior perfectly fits all the assumptions of the kinetic-molecular theory. In reality, gases are not ideal, but are very close to being so under most everyday conditions.

The kinetic-molecular theory as it applies to gases has five basic assumptions.

  • Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size.
  • Gas particles are in constant rapid motion in random directions.
  • Collisions between gas particles and between particles and the container walls are elastic collisions.
  • The average kinetic energy of gas particles is dependent upon the temperature of the gas.
  • There are no forces of attraction or repulsion between gas particles.
5 0
3 years ago
A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula
Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

= 2.857 x 10^14 g of gold in the ocean.

1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

= $5,840,000,000,000,000

8 0
3 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
What would happen to the block if it had a density of 0.500 kg/L and was placed in the same 100.0 L tank of water?
raketka [301]
I’m not sure if there was important information in the question before this one, but the answer based on the info I have is B.

The density of water is 1kg/L. Since the density of the block is less, it will float.
6 0
3 years ago
Read 2 more answers
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