I NEED HELP ASAP PLEASEEE

Wittaler [7]

Gas particles are in constant motion, and any object in motion has kinetic energy (Ek). ... For example, in the collision of two molecules, one molecule may be deflected at a slightly higher speed and the other at a slightly lower speed, but the average kinetic energy does not change.
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3 0

3 years ago

If ice is warmed and becomes a liquid, which type of process is it ?

attashe74 [19]

The answer is Endothermic

6 0

3 years ago

Read 2 more answers

Write the full ground state electron configuration of f+.

ANTONII [103]

Answer:- $1s^22s^22p^4$ .

Explanation:- Atomic number for fluorine(F) is 9 and it's electron configuration is $1s^22s^22p^5$ . $F^+$ is formed when F loses one electron from it's valence shell.

$F\rightarrow F^++1e^-$

Second shell is the valence shell for fluorine and so it loses one electron from 2p to form $F^+$ and the electron configuration of the ion becomes $1s^22s^22p^4$ .

7 0

4 years ago

Calculate the number of grams contained in each of the following number of moles. Report your answer with the appropriate number

Debora [2.8K]

Answers:

a. 131.85 grams

b. 134.69 grams

To calculate the number of grams in the number of moles given for each compound, consider the formula:

Moles = No. of grams / molar mass

Therefore, to find the gram it will become:

No. of grams = moles x molar mass

a. 0.410 mol of NaCl (sodium chloride)

Molar mass of NaCl = (286.14 + 35.45) = 321.59

No. of grams = moles x molar mass

No. of grams = 0.410 x 321.59 = 131.85 grams

b. 4.80 mol of C2H4 (ethylene)

Molar mass of C2H4 = (12.01 x 2) + (1.01 x 4) = 28.06

No. of grams = moles x molar mass

No. of grams = 4.80 x 28.06 = 134.69 grams

3 0

3 years ago

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Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$

But this is just a supposition.

Regards.

4 0

3 years ago