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1 year ago

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a compound contain 8.57g of carbon and 1.43g of hydrogen. The relative formula mass is 70 calculate the empirical formula of the

compound and the molecular formula of the compound

1 answer:

yanalaym [24]1 year ago

5 0

Answer:

Empirical formula: CH2

Molecular formula: C5H10

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Calculate the mass of MNO2 needed to produce 25.0 g of Cl2

Vedmedyk [2.9K]

I will solve this question assuming the reaction equation look like this:
<span>MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O.
</span>
For every one molecule of MnO2 used, there will be one molecule of Cl2 formed. If the molecular mass of MnO2 is 87g/mol and molecular mass of Cl2 is <span> 73.0 g/mol, the mass of MnO2 needed would be:
Cl mass/Cl molecular mass * MnO2 molecular mass=
25g/ (73g/mol) * (87g/mol) * 1/1= 29.8 grams</span>

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2 years ago

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You have a mixture that contains 0.380 moles of Ne(g), 0.250 moles of He(g), and 0.500 moles CH4(g) at 400 K and 7.25 atm. What

boyakko [2]

I think it is the pacific ocean

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2 years ago

a compound contain 8.57g of carbon and 1.43g of hydrogen. The relative formula mass is 70 calculate the empirical formula of the

yanalaym [24]

Answer:

Empirical formula: CH2

Molecular formula: C5H10

5 0

1 year ago

9) After lab, all of Darrel’s friends looked at his data and laughed and laughed. They told him that he was 30.8% too low in the

zaharov [31]

If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.

In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.

So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC

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2 years ago

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What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)

blsea [12.9K]

<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, $(C_{2}H_{5})_{3}N$ , the reaction will be
$(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}$

and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value = 5.3 x 10^-4

And $kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}$

thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

=0.00516 M

Thus, pOH = 2.30
pH = 14 - pOH = 11.7

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2 years ago