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Aneli [31]
3 years ago
11

Need helpppp please

Chemistry
1 answer:
Vladimir79 [104]3 years ago
6 0

Answer:

message me so i can help you because i cant see the words

Explanation:

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calculate the volume of carbon dioxide of room temperature and pressure obtained from 30 grams of glucose​
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Answer:

A

Explanation:

4 0
3 years ago
For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.
Charra [1.4K]

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3 0
2 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>

3 0
3 years ago
Read 2 more answers
Does anyone know how to fill in this diagram?
tangare [24]

Answer:

solid to gas is sublimation

gas to solid is deposition

liquid to gas is evaporation/boiling

the rest are correct

3 0
3 years ago
Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.
sergejj [24]

The molecular formula of compound is Al(C_{2}H_{3}O_{2})_{3}.

Since, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6.023\times 10^{23} atoms of  Al(C_{2}H_{3}O_{2})_{3}.

Thus, according to molecular formula, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6\times 6.023\times 10^{23}=3.61\times 10^{24} atoms of oxygen atoms.

1 atom of oxygen will be present in \frac{1}{3.61\times 10^{24}} moles of Al(C_{2}H_{3}O_{2})_{3} . Thus,

2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}.

Molar mass of Al(C_{2}H_{3}O_{2})_{3} is 236 g/mol, mass can be calculated as follows:

m=n\times M=0.7285 mol\times 236 g/mol=172 g

Therefore, mass of Al(C_{2}H_{3}O_{2})_{3} will be 172 g.

5 0
3 years ago
Read 2 more answers
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