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Rashid [163]
3 years ago
12

If 35.4 liters of hydrogen gas are used, how many liters of nitrogen gas will be needed for the above reaction at STP?

Chemistry
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

ujdbdbfbf fbFbsygtsjysnsgndhngsnsgnafbfa

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What is the oxidation half-reaction for 2Mg + O2 + 2Mgo?
miv72 [106K]

Answer:

B => Mg° => Mg⁺² + 2e⁻

Explanation:

2Mg° + O₂ => 2MgO  rewrite => Mg° + 1/2O₂ => 2MgO

Oxidation Half Rxn => Mg° => Mg⁺² + 2e⁻

Reduction Half Rxn => 1/2O₂ + 2e⁻ => 2O⁻

__________________________________-

FYI Note =>  For half-reactions one always finds the electrons listed on the product side of the reaction for oxidation reactions and for reduction reactions on the reactant side.

5 0
3 years ago
100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.
yawa3891 [41]

Volume of H₂O added = 175 ml

<h3>Further explanation</h3>

Given

100 gm of a 55% (M/M)  and 20% (M/M) nitric acid solution

Required

waters added

Solution

starting solution

mass H₂O = 45%=45 g

%mass of H₂O in new solution = 100%-20%=80%

Can be formulated for %mass H₂O :

\tt \dfrac{45+x}{100+x}=80\%\\\\45+x=0.8(100+x)\\\\45+x=80+0.8x\\\\35=0.2x\rightarrow x=175~g

For water mass=volume(density = 1 g/ml)

So volume added = 175 ml

7 0
3 years ago
Which of the following is an example of organic insecticides? a) Aldrin b) Lead arsenate c) Lime sulphur d) Fluorides​
PolarNik [594]

Explanation:

i chose b, but im not so sure.

6 0
2 years ago
What is the volume of an 2.3 solution with 212 grams of calcium chloride dissolved in it
Travka [436]

830 mL. A 2.3 mol/L solution of CaCl2 has a volume of 830 mL

I am guessing that the concentration of your solution is 2.3 mol/L.

a) Moles of CaCl2

MM of CaCl2 = 110.98 g/mol

Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)

= 1.910 mol CaCl2

b) Volume of solution

V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution

= 830 mL solution


5 0
3 years ago
Which one of the following equations represents the net ionic equation for the reaction between aqueous potassium chloride and a
Nana76 [90]
<h3>Answer:</h3>

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

<h3>Explanation:</h3>

The questions requires we write the net ionic equation for the reaction between aqueous potassium chloride and aqueous silver nitrate.

<h3>Step 1: Writing a balanced equation for the reaction.</h3>
  • The balanced equation for the reaction between aqueous potassium chloride and aqueous silver nitrate will be given by;

KCl(aq) + AgNO₃(aq) → KNO₃(aq) +AgCl(s)

  • AgCl is the precipitate formed by the reaction.
<h3>Step 2: Write the complete ionic equation.</h3>
  • The complete ionic equation for the reaction is given by showing all the ions involved in the reaction.

K⁺(aq)Cl⁻(aq) + Ag⁺(aq)NO₃⁻(aq) → K⁺(aq)NO₃⁻(aq) +AgCl(s)

  • Only ionic compounds are split into ions.
<h3>Step 3: Write the net ionic equation for the reaction.</h3>
  • The net ionic equation for a reactions only the ions that fully participated in the reaction and omits the ions that did not participate in the reaction.
  • The ions that are not involved directly in the reaction are known as spectator ions and are not included while writing net ionic equation.

Ag⁺(aq) +Cl⁻(aq) → AgCl(s)

4 0
3 years ago
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