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Korvikt [17]
3 years ago
9

I need help on this question please help

Chemistry
1 answer:
sleet_krkn [62]3 years ago
3 0

Answer:

2H₂ + O₂ —> 2H₂O

The coefficients are: 2, 1, 2

Explanation:

__ H₂ + __ O₂ —> __ H₂O

The above equation can be balance as illustrated below:

H₂ + O₂ —> H₂O

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by writing 2 before H₂O as shown below:

H₂ + O₂ —> 2H₂O

There are 2 atoms of H on the left side and 4 atoms on the right side. It can be balance by writing 2 before H₂ as shown below:

2H₂ + O₂ —> 2H₂O

Now, the equation is balanced and the coefficients are: 2, 1, 2

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The oxidation state of phosphorus is +3 in
IgorC [24]

Answer:

b) Phosphorus acid

Explanation:

To distinguish the type of acid of phosphorus with the oxidation state of +3, we need to be familiar with the chemical formula of each of the compounds:

    Orthophosphoric acid             H₃PO₄

    Phosphorus acid                       H₃PO₃

    Metaphosphoric acid               HPO₃

    Phyrophosphoric acid​               H₄P₂O₇

Now that we know the formula of the given compounds, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero:

Only phosphorus acid yielded an oxidation state of +3 for phosphorus in the compound.

  H₃PO₃:

   we know the oxidation state of H = +1

                                                          O = -2

         The oxidation state of P is unknown. We can express this as an equation:

                3(+1) + P + 3(-2) = 0

                    3 + P -6 = 0

                          P-3 = 0

                          P = +3

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3 years ago
Mengapa garam NaNO2 dapat membirukan kertas lakmus merah? Jelaskan dengan reaksi kimianya!
Reika [66]
Can't understand this. English Please!  :) :)
8 0
3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
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