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2 years ago

Which description best characterizes the motion of particles in a solid

2 answers:

8 0

<span>The movement of particles within a solid is extremely slow when compared to that of a gas. It is also significantly slower in movement than that of the movement found within the particles of liquid. The more movement present and the faster the movement of the particles the more space will be present between each particle. This causes the material to spread out as they become less densely packed within a solid material.</span>

11111nata11111 [884]2 years ago

3 0

Answer:

Vibrating around fix positions

Explanation:

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PLEASE ANSWER I AM BEGGING

TiliK225 [7]

Taking into account the definition of dilution:

you have to use 8.23 mL of a stock solution of 7.00 M HNO₃ to prepare 0.120 L of 0.480 M HNO₃.
If you dilute 20.0 mL of the stock solution to a final volume of 0.270 L , the concentration of the diluted solution is 0.518 M.

<h3>Dilution</h3>

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

Ci: initial concentration
Vi: initial volume
Cf: final concentration
Vf: final volume

<h3>Volume of stock solution</h3>

In this case, you know:

Ci= 7 M
Vi= ?
Cf= 0.480 M
Vf= 0.120 L

Replacing in the definition of dilution:

7 M× Vi= 0.480 M× 0.120 L

Solving:

Vi= (0.480 M× 0.120 L)÷ 7 M

<u><em>Vi= 0.00823 L= 8.23 mL</em></u> (being 1 L= 1000 mL)

Finally, you will need 8.23 mL of the stock solution.

<h3>Concentration of the diluted solution</h3>

In this case, you know:

Ci= 7 M assuming the stock solution is 7.00 M HNO₃
Vi= 20 mL= 0.02 L
Cf= ?
Vf= 0.270 L

Replacing in the definition of dilution:

7 M× 0.02 L= Cf× 0.270 L

Solving:

(7 M× 0.02 L)÷ 0.270 L= Cf

Finally, the concentration is 0.518 M.

Learn more about dilution:

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#SPJ1

5 0

1 year ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago

Short, hairlike appendage

LekaFEV [45]

Cilium part of a cell

6 0

3 years ago

Ochem question?

insens350 [35]

Answer: 4.) CH3O is the answer, which stands for Methoxide.

Explanation:

3 0

2 years ago

Calculate the mass of one mole of these substances.

pogonyaev

Answer:

a. 53.5 g/mol

b. 80.06 g/mol

c. 133.33 g/mol

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

Reading a Periodic Table
Molar Mass - 1 mol per <em>x</em> grams substance

Explanation:

<u>Step 1: Define</u>

a. NH₄Cl

b. NH₄NO₃

c. AlCl₃

<u>Step 2: Find masses</u>

Molar Mass of N - 14.01 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Al - 26.98 g/mol

Molar Mass of Cl - 35.45 g/mol

<u>Step 3: Calculate compound masses</u>

Molar Mass of NH₄Cl - 14.01 g/mol + 4(1.01 g/mol) + 35.45 g/mol = 53.5 g/mol

Molar Mass of NH₄NO₃ - 2(14.01 g/mol) + 4(1.01 g/mol) + 3(16.00 g/mol) = 80.06 g/mol

Molar Mass of AlCl₃ - 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol

6 0

3 years ago