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const2013 [10]
3 years ago
11

Na + O2 → Na2O2 Can you guys help me !

Chemistry
1 answer:
vfiekz [6]3 years ago
6 0

Answer:

Explanation:

Assuming that you are asking for the full-balanced equations...

2Na + O2 = Na2O2

H2 + O2 = H2O2; This equation is already balanced, there is no need for change.  

Please let me know if this is not the answer you are looking for and I will amend the answer.

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Air should be classified as:
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Explanation:

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What is the molar mass of (NH4)3 PO3?
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133.0873 g/mol

(NH4)3PO3 - molar mass
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You wish to make a 0.285 M hydroiodic acid solution from a stock solution of 12.0 M hydroiodic acid. How much concentrated acid
Inessa05 [86]

The amount, in mL, of the concentrated acid required, would be 1.1875 mL

<h3>Dilution</h3>

From the dilution equation:

m1v1=m2v2 where m1 and m2 = molarity before and after dilution, and v1 and v2 = volume before and after dilution.

m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL

v1 = m2v2/m1 = 0.285x50/12 = 1.1875 mL

Thus, 1.1875 mL of the acid would be taken and diluted with water up to the 50 mL mark.

More on dilution can be found here: brainly.com/question/13949222

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2 years ago
Phosphine, PH3(g), decomposes according to the equation 4PH3(g) --&gt; P4(g) + 6H2(g) The kinetics of the decomposition of phosp
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PLEASE HELP!!!!!!!
dsp73

Answer: 1. 2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu

2. 3 moles of CuCl_2 : 2 moles of Al

3. 0.33 moles of CuCl_2 : 0.92 moles of Al

4. CuCl_2 is the limiting reagent and Al is the excess reagent.

5. Theoretical yield of AlCl_3 is 29.3 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al=\frac{25.0g}{27g/mol}=0.92moles

\text{Moles of} CuCl_2=\frac{45.0g}{134g/mol}=0.33moles

The balanced chemical equation is:

2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu  

According to stoichiometry :

3 moles of CuCl_2 require = 2 moles of Al

Thus 0.33 moles of CuCl_2 will require=\frac{2}{3}\times 0.33=0.22moles  of Al

Thus CuCl_2 is the limiting reagent as it limits the formation of product and Al is the excess reagent.

As 3 moles of CuCl_2 give = 2 moles of AlCl_3

Thus 0.33 moles of CuCl_2 give =\frac{2}{3}\times 0.33=0.22moles  of AlCl_3

Theoretical yield of AlCl_3=moles\times {\text {Molar mass}}=0.22moles\times 133.34g/mol=29.3

Thus 29.3 g of aluminium chloride is formed.

5 0
3 years ago
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