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geniusboy [140]
4 years ago
7

What is the molar mass of helium? (show your work)

Chemistry
1 answer:
jeka57 [31]4 years ago
5 0

Answer:

4.002602 u

Explanation:

Idk how to show my work.

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Equation of ethane to ethanoic acid
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Feb 16, 2012 ... How do you convert ethane to ethanoic acid? (with equation please). First you chlorinate it in presence of light. C2H6 + Cl2 ---hv -> C2H5Cl + ...

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he formation of the aldehyde is shown by the simplified equation: ... Note: The equation for the conversion of ethanol to ethanoic acid is worked out in detail ... Starting from ethanenitrile, you would therefore get a solution containing ethanoate ... Mar 3, 2012 ... As well as using a normal type of molecular formula to describe an organic .... In ethane the carbon atoms have the maximum number of hydrogen atoms bonded .... e.g. ethanoic acid and ethanol will produce ethyl ethanoate,.


8 0
3 years ago
What is the molarity of a solution made by dissolving 1.25 mol of HCl in enough
Ronch [10]

Answer:

Explanation:

C

4 0
3 years ago
Scientists are attempting to determine the composition of an unknown compound. The compound has a high melting point and is a go
zlopas [31]

Answer C

Explanation:

I took the test

3 0
3 years ago
Which solution has the same boiling point as 0.25 mol CaCl2 dissolved in 1000 g water?
Rufina [12.5K]
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality). 

 ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.

5 0
3 years ago
in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc
Margarita [4]

Answer:  0.1161 grams of mercury(II) sulfide) form.

Explanation:

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 L)}}     .....(1)

a) Molarity of Na_2S solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol

\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol

b) Molarity of Hg(NO_3)_2 solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol

Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3

According to stoichiometry :

1 mole of Hg(NO_3)_2 reacts with 1 mole of Na_2S

Thus 0.0005 moles of HgNO_3 reacts with=\frac{1}{1}\times 0.0005=0.0005 moles of Hg(NO_3)_2

Thus Hg(NO_3)_2 is the limiting reagent and Na_2S is the excess reagent.

According to stoichiometry :

1 mole of Hg(NO_3)_2 forms=  1 mole of Hg_2S

Thus 0.0005 moles of Hg(NO_3)_2 forms=\frac{1}{1}\times 0.0005=0.0005 moles of Hg_2S

mass of H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0
3 years ago
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