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3 years ago

Emission nebulae are also called ____ because they are composed of ionized hydrogen.

1 answer:

8 0

H II regions. The ionized gas emits various colors of light and creates some of the most beautiful formations in the observable universe. These are slo regions where star formation is often taking plan and young stars are being formed from the remnants of older stars.

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Calculate the molarity of 29.9 g of MgS in 911 mL of solution??

LUCKY_DIMON [66]

0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Define the molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Given data:

Volume of solution = 911 mL=0.911L

Given the mass of MgS = 29.9 g

Molar mass of MgS = 56.38 g/mol

Moles of MgS = 0.5303299042

$Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}$

$Molality = \frac{0.5303299042}{0.911L}$

Molality = 0.5821403998=0.58

Hence, 0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Learn more about Molarity here:

brainly.com/question/2817451

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3 0

2 years ago

Calculate the percent ionization of a 0.15 M benzoic acid solution in pure water and in a solution containing 0.10 M sodium benz

Hoochie [10]

Answer:

% ionization for benzoic acid = 0.08%

% ionization for sodium benzoate = 2.5%

The percentage ionization differ significantly because benzoic acid is a weak acid while sodium benzoate is a salt of benzoic acid. Their extent of dissociation also differ because they were compared in different solutions

Explanation:

Ka for pure water = 1.0 * 10-⁷

Ka for sodium benzoate = 6.5*10-⁵

1. For benzoic acid (C6H5COOH)

C6H5COOH ==== C6H5COO‐ + H+

0.15M 0 0

0.15-x x x

Ka = [C6H5COO-] [H+] / [C6H5COOH]

Ka = [X] [X] / 0.15 - X

1.0*10-⁷ = [X]² / 0.15 - x

But x is negligible compared to 0.15,

(1.0*10-⁷)*0.15 = x²

Take square root of both sides,

X = 1.22 * 10-⁴

% ionization = ( [H+] / [C6H5COOH] ) * 100

% ionization = (1.22*10-⁷ / 0.15) * 100

% ionization = 0.08%

2. For C6H5COONa

Note: I will not repeat the same procedure of dissociation again since they're basically the same just the difference in ions

Ka for C6H5COONa = 6.5*10-⁵

6.5*10-⁵ = [X]² / (0.10 - X)

Cross-multiply both sides;

(6.5*10-⁵ * 0.10) = X²

Take square root of both side,

X= 2.5*10-³

% ionization = (2.5*10-³ / 0.10) *100

% ionization = 2.5%

5 0

3 years ago

What did Rutherford’s scattering experiment reveal about the structure of atoms?

Stels [109]

Answer:

Explanation:

This experiment by Rutherford involved the firing of alpha particles at gold foils. It is also. called the gold foil experiment.

He fired these alpha particles at different points. He noticed that at some points, there were deflections, while at some other points, there were no deflections. It is necessary to state that these alpha particles are positively charged. For there to be a deflection, there must have been a kind of repulsion between the gold foil and the alpha particles.

From the basic physics of like repels like, he knew for sure that there must be dense positive core in the atom that is causing the deflection of the alpha particles. This enabled him to come up with the theory that the atom contained a small dense positive core called the nucleus

7 0

4 years ago

What is the term for a solid that forms when two liquids are mixed together

MrRissso [65]

The correct answer would be C. a precipitate
Hope this helped!

5 0

3 years ago

Read 2 more answers

Given the following data, determine the order of reaction with respect to H2.

alex41 [277]

Answer:

First-order with respect to hydrogen.

Explanation:

Hello!

In this case, considering that the rate law of this reaction can be expressed via:

$r=k*p_{H_2}^m*p_{ICl}^n$

If we want to know m, the order of reaction with respect to hydrogen, we need to relate the experiments 1 and 3 in order to get rid of the pressure of ICl:

$\frac{r_1}{r_3} =\frac{k*p_{H_2,1}^m*p_{ICl,1}^n}{k*p_{H_2,3}^m*p_{ICl,3}^n}$