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saw5 [17]
3 years ago
5

What is the molar mass of O2 A. 16 g B. 16 AMU C. 32 AMU D. 32 g

Chemistry
2 answers:
Alexeev081 [22]3 years ago
5 0
The answer is D. 32G

MArishka [77]3 years ago
3 0
D.32 g will be your answer hope that helps
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Write the full ground state electron configuration of C − C− . electron configuration: Which neutral atom is isoelectronic with
stepan [7]

Answer: 1. Is2 2s2 2p3

2. Nitrogen

Explanation: The number of electron present In C = 6

But an extra electron is added since the charge on C is -1, this therefore makes the total electron 7.

1. By arrangement, the Electronic configuration is therefore;

Ans: 1s2 2s2 2p3

2. It is explained how C has 7 electrons, we can proceed then.

Neutral atom have atomic number of 7.

The element with atomic number of 7 is;

Ans: NITROGEN

7 0
3 years ago
How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
ahrayia [7]
<span>Answer: 100 ml
</span>

<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>

iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>

<span>2) Use the percent yield to calculate the theoretical amount:
</span>

<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
</span>

<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
</span>

<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
</span>

<span>i) Molarity formula: M = n / V
</span>

<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>

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Fine the volume in dm³ and in Ml of a 0.505 mol/dm of NaCH required to react with 40Ml of 0.505 molar concentration of H2SO4​
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