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2 years ago

What volume in L of a 0.724 M Nal solution contains 0.405 mol of Nal?

Chemistry

2 answers:

bekas [8.4K]2 years ago

6 0

Answer:

Explanation:

I added the pictures for anyone struggling with that aspect

Setler [38]2 years ago

4 0

Answer:

0.559 L

Explanation:

Step 1: Given data

Moles of sodium iodide (n): 0.405 mol

Molar concentration of sodium iodide (M): 0.724 M (0.724 mol/L)

Step 2: Calculate the volume of solution (V)

The molarity is equal to the moles of solute divided by the liters of solution.

M = n/V

V = n/M

V = 0.405 mol/(0.724 mol/L) = 0.559 L

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Role and significance of guassion or normal distribution?

kkurt [141]

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2 years ago

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A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

RoseWind [281]

Answer:

$C_3=0.125M$

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

$V_1C_1=V_2C_2$

So we solve for C2:

$C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M$

2. Now, since 111 mL of water is added, we compute the final volume, V3:

$V_3=139+111=250mL$

So, the final concentration of the 139 mL portion is:

$C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M$

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8 0

2 years ago

What happens to the temperature and density of the material between points B and C?

makvit [3.9K]

Answer:

Temperature decreases and density increases

Explanation:

Let us remember that density of a material increases as the temperature of the material decreases. So the cooler a material becomes, the denser it becomes also.

Between points B and C, the material rapidly cools down and the temperature decreases accordingly. This ultimately results in an increase in density since cooler materials are denser than hot materials.

7 0

2 years ago

Explain specifically how an electron gives off light in an atom.

exis [7]

Answer:

Then, at some point, these higher energy electrons give up their "extra" energy in the form of a photon of light, and fall back down to their original energy level.

Explanation:

When properly stimulated, electrons in these materials move from a lower level of energy up to a higher level of energy and occupy a different orbital.

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2 years ago

Describe Rutherford's contribution to the atomic model.

Nady [450]

Answer:

Rutherford's experiment, also known as

$\alpha - scattering \: experiment$

supports the existence of neutrons and the nucleus.

Explanation:

In the above diagram, Rutherford was trying to explain his contributions using thin foils of gold and other metals as targets for alpha particles from a radioactive source.

He observed that the majority of particles penetrated the foil either undeflected or with only a slight deflection. But, every now and then an alpha particle was scattered(or deflected) at a large angle..

According to Rutherford, most of the atoms must be empty space. This explains why the majority of alpha particles passed through through the gold foil with little or no deflection. The atoms positive charges, Rutherford proposed are all concentrated in the Nucleus, which is a dense central core within the atom.

Whenever an alpha particle came close to a nucleus in the scattering experiment, it experienced a large repulsive force and therefore a large deflection. Moreover, an alpha particle coming towards a nucleus would be completely repelled and its direction would be reversed. The positively charged particles in the Nucleus are called Protons.

I hope you find this useful... Have a lovely day. 

4 0

2 years ago