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laila [671]
3 years ago
9

when 0.72 g of a liquid is vaporized at 110° C and 0.967 atm, the gas occupies a volume of 0.559L. The empirical formula of the

gas is CH2. what is the molecular formula of the gas?
Chemistry
1 answer:
Leno4ka [110]3 years ago
8 0
Hydrogen gas mixed with sulfur
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Weak acids
maria [59]

Answer:

congratulations  you have made cocaine

3 0
3 years ago
What two quantities must be known to calculate the density of a sample of matter? A) mass and volume B) length and mass C) size
Svetradugi [14.3K]

Answer:

A

Explanation:

4 0
3 years ago
Read 2 more answers
Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338
Elden [556K]

<u>Answer:</u> The concentration of chloride ions in the solution obtained is 0.674 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

  • <u>For KCl:</u>

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

  • <u>For magnesium chloride:</u>

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = (2\times 0.224)=0.448mol

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

5 0
3 years ago
Complete the following single replacement reaction. If they don’t react, just write “NR”
Kipish [7]

Here we have to complete the given single replacement reactions.

The replacement reactions are-

1) Fe (s) + CuCl₂ (aq) → FeCl₂ (aq) + Cu (s)

2) Cu (s) + FeCl₂ (aq) → NA

3) K (s) + NiBr₂ (aq) → NA

4) Ni (s) + KBr (aq) → NiBr₂ (aq) + K (s)

5) Zn (s) + Ca(NO₃)₂ (aq) → Zn(NO₃)₂ (aq)  + Ca (s)

6) Ca (s) + Zn(NO₃)₂ (aq) → NA

The replacement reactions can be explained in light of the redox potential.

The standard reduction potential of the half cells involved in these reactions are:

Fe²⁺ + 2e → Fe (E° = -0.441V); Cu²⁺ + 2e → Cu (E° = 0.674V)

Ni²⁺ + 2e → Ni (E° = -0.23V); Zn²⁺ + 2e → Zn (E° = -0.763V)

We know the half cell reactions in which the standard reduction potentials are positive are allowed.

1) The reaction is possible as Cu²⁺/Cu and Fe/Fe²⁺ standard reduction potentials are positive.

2) The reaction is not possible as Cu/Cu²⁺ and Fe²⁺/Fe standard reduction potentials are negative.

3) The reaction is not possible as Ni²⁺/Ni standard reduction potential is negative.

4) The reaction is possible as Ni/Ni²⁺ standard reduction potential is positive.

5) The reaction is possible as Zn/Zn²⁺ standard reduction potential is positive.

6) The reaction is possible as Zn²⁺/Zn standard reduction potential is negative.

4 0
2 years ago
Answer this question pleas
Korvikt [17]

Answer:nope

Explanation:

Just no

4 0
3 years ago
Read 2 more answers
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