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Svetlanka [38]
2 years ago
15

1. How many Chromium atoms are found in 25.8 milligrams of Chromium?​

Chemistry
2 answers:
solong [7]2 years ago
8 0

Answer:

Chromium is a steel-gray, lustrous, hard metal that takes a high polish, is fusible with difficulty, and is resistant to corrosion and tarnishing.

The most common oxidation states of chromium are +2, +3, and +6, with +3 being the most stable. +4 and +5 are relatively rare. Chromium compounds of oxidation state 6 are powerful oxidants.ion:

givi [52]2 years ago
7 0

Answer: 51.9961 g/mol, don't know if it helps :)

Explanation:

You might be interested in
Identify one disadvantage to each of the following models of electron configuration:
Murrr4er [49]

Answer:

The disadvantages of each of the given model of electron configuration have been mentioned below:

1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.

2). Arrow and line diagrams make the counting of electrons and take up too much space.

3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.

6 0
1 year ago
Read 2 more answers
Use the PhET simulation to identify what happens to the concentrations of a dilute Drink mix mixture or a pure Drink mix solutio
elena-s [515]

Hello user, you did not add a picture of an attachment to enable me help you solve this problem. Please check the attachment I added i believe it is the question that needs to be solved.

Explanation:

1. In the first column

<u>INCREASE CONCENTRATION</u>: these are the items that needs to be placed in this bin

a. Add drink mix solid to a diluted mixture of drink mix in pure water

b. Add drink mix solution to a diluted mixture of drink mix in pure water

c. Add drink mix solid to pure drink mix solution

d Evaporate water from a diluted mixture of drink mix in pure water

e Evaporate water from pure drink mix solution

2. The items below are what should be placed in this bin,<u>DECREASE CONCENTRATION:</u>

a. Add water to a diluted mixture of drink mix in pure water

b. Add water to pure drink mix solution

3. These items here should be placed here. <u>DOES NOT AFFECT CONCENTRATION:</u>

a. Add pure drink mix solution to pure drink mix solution

b. Drain the pure drink mix solution

c. Drain the diluted mixture

7 0
2 years ago
A solution was prepared by mixing 50.0 g
frez [133]

Answer:

4.78 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em></em>

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>

4 0
2 years ago
The bond dissociation energy to break 4 bond(s) in 1 mole of CH₄ molecules is:_____ **Any help would be greatly appreciated!**
frozen [14]

Answer:

The bond dissociation energy to break 4 bonds in 1 mol of CH is 1644 kJ

Explanation:

Since there are 4 C-H bonds in CH₄, the bond dissociation energy of 1 mol of CH₄ is 4 × bond dissociation energy of one C-H bond.

From the table one mole is C-H bond requires 411 kJ, that is 411 kJ/mol. Therefore, 4 C-H bonds would require 4 × 411 kJ = 1644 kJ

So, the bond dissociation energy to break 4 bonds in 1 mol of CH₄ is 1644 kJ

5 0
2 years ago
Given that the rate constant is 4.0×10−4 m−1 s−1 at 25.0 ∘c and that the rate constant is 2.6×10−3 m−1 s−1 at 42.4 ∘c, what is t
stepladder [879]
So here's how you find the answer:

Given: (rate constants)

K₁ = 4.0 x 10⁻⁴ M⁻¹s⁻¹.
T₁ = 25.0 C = 293 K.
k₂ = 2.6 x10⁻³ M⁻¹s⁻¹.
T₂ = 42.4 C = 315.4 K.
R = 8,314 J/K·mol.

Use the equation:

ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).

Transpose:

Ea = R·T₁·T₂ / (T₁ - T₂) · ln(k₁/k₂)

Substitute within the given transposed equation:

<span>Ea = 8,314 J/K·mol · 293 K · 315.4 K ÷ (293 K - 315.4 K) · ln (4.0 x 10⁻⁴ M⁻¹s⁻¹/ 2.6 x 10⁻³ M⁻¹s⁻¹).
</span>
Continuing the solution we get:

<span>Ea = 768315 J·K/mol ÷ (-22,4 K) * (-1.87)
</span>
The value of EA is:

<span>Ea = 64140.58 J/mol ÷ 1000 J/kJ = 64.140 kJ/mol.</span>


4 0
2 years ago
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