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pickupchik [31]
3 years ago
6

b. Determine the molecular formula of a molecule with an empirical formula NH2and a molar mass of 32 g/mol.

Chemistry
1 answer:
VladimirAG [237]3 years ago
5 0
I'm taking this lesson now, so imma help u ( if u need anything else ask me)

so given Molar mass= 32 g/mol
molar mass= (empirical formula) n
32 = (14x1 + 2x1) n
32 = 16 n , so n= 2
so, molecular formula= N2H4
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Yakvenalex [24]

Answer:

T_2=-125.58\°C

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Whereas solving for the final temperature T2, we get:

T_2=\frac{T_1P_2}{P_1}

Thus, we plug in the given data (temperature in Kelvins) to obtain:

T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C

Best regards!

3 0
2 years ago
You are given a solid that is a mixture of na2so4 and k2so4.
Murljashka [212]

Here we have to calculate the amount of SO_{4}^{2-} ion present in the sample.

In the sample solution 0.122g of SO_{4}^{2-} ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ +  SO_{4}^{2-}

(Na)₂SO₄=2Na⁺ +  SO_{4}^{2-}

Thus, BaCl₂+  SO_{4}^{2-} = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and  K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of  SO_{4}^{2-} ion is precipitated in this reaction.  

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion (SO_{4}^{2-}) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of SO_{4}^{2-} ion is present. So in 1 g of BaSO₄ \frac{96.06}{233.3}=0.411 g of SO_{4}^{2-} ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of SO_{4}^{2-} ion is present.        

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Hi dear
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