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Karolina [17]
3 years ago
15

What mass will 33.6 l of chlorine gas (cl2) have at stp

Chemistry
1 answer:
Aloiza [94]3 years ago
5 0
The STP stands for standard temperature and pressure which means in a condition where the temperature is 273.15K and the pressure is 0.986atm. In STP, an ideal gas will have a volume 22.4 liters for every mol of gas. Then, the amount of molecule of the <span>33.6 l of chlorine gas (Cl2) would be:

volume of gas/ number of mol=  22.4l/mol
</span>33.6l/ number of mol=  22.4l/mol
<span>number of mol= 33.6l / (22.4 liters/mol)= 1.5 moles

The mass of the gas would be: 1.5 moles * </span><span>70.906 </span>g/mol= 106.359 grams
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How would you prepare 500 mL of 0.360 M solution of CaCl2 from<br> solid CaCl2?
LenKa [72]

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

0.500 L \times \frac{0.360mol}{L} = 0.180 mol

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

0.180 mol \times \frac{110.98g}{mol} = 20.0 g

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.

We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.

You can learn more about solutions here: brainly.com/question/2412491

4 0
2 years ago
A cyanide solution with avolume of 12.73 mL was treated with 25.00 mL of Ni2+solution (containing
natima [27]

Answer:

The molarity of this solution is 0,09254M

Explanation:

The concentration of the Ni²⁺ solution is:

Ni²⁺ + EDTA⁴⁻ → Ni(EDTA)²⁻

0,03935L × 0,01307M = 5,143x10⁻⁴ moles Ni²⁺ ÷ 0,03010L =<em>0,01709M Ni²⁺</em>

25,00 mL of this solution contain:

0,01709M × 0,02500L = 4,2716x10⁻⁴ moles of Ni²⁺

The moles of Ni²⁺ that are in excess and react with EDTA⁴⁻ are:

0,01015L × 0,01307M = 1,3266x10⁻⁴ moles of Ni²⁺

Thus, moles of Ni²⁺ that react with CN⁻ are:

4,2716x10⁻⁴ - 1,3266x10⁻⁴ = 2,9450x10⁻⁴ moles of Ni²⁺

For the reaction:

4CN⁻ + Ni²⁺ → Ni(CN)₄²⁻

Four moles of CN⁻ react with 1 mole of Ni²⁺:

2,9450x10⁻⁴ moles of Ni²⁺ × \frac{4 mol CN^-}{1 molNi^{2+}} = <em>1,178x10⁻³ moles of CN⁻</em>

As the volume of cyanide solution is 12,73mL. The molarity of this solution is:

<em>1,178x10⁻³ moles of CN⁻ ÷ 0,01273L = </em><em>0,09254M</em>

I hope it helps!

5 0
3 years ago
Gerald's science teacher mixed liquid X and liquid Y, both at room-temperature, in a large beaker. two beakers of clear solution
Pachacha [2.7K]

Answer:

the answer is C

Explanation:

7 0
3 years ago
Read 2 more answers
26. Balance the following equations:<br> Ca(s) + H3PO4(aq)Ca3(PO4)2(s) + H2(g)
yKpoI14uk [10]

Hey there!

Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂

Balance PO₄.

1 on the left, 2 on the right. Add a coefficient of 2 in front of H₃PO₄.

Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂

Balance H.

6 on the left, 2 on the right. Add a coefficient of 3 in front of H₂.

Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Balance Ca.

1 on the right, 3 on the right. Add a coefficient of 3 in front of Ca.

3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Our final balanced equation:

3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Hope this helps!

3 0
3 years ago
Chemical properties of sodium
Svetlanka [38]

Answer:

Density 0.97 g.cm -3 at 20 °C

Melting point 97.5 °C

Boiling point 883 °C

oxidation states +1, −1 (rare)

Explanation:

Have a nice day

5 0
3 years ago
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