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SVETLANKA909090 [29]
3 years ago
5

What is the mass percent of sodium in NaCl? a)39.34% b)60.66% c)60.66% d)78.68%

Chemistry
2 answers:
rosijanka [135]3 years ago
6 0
Well as it is 39.33% then go with a. also with 58.44 grams of NaCl you have 22.99 g Na so x100% = 39.33%
vodka [1.7K]3 years ago
3 0

Answer:

Explanation:

Molecular mass of NaCl = 58.44g/mol

Molecular mass of Na = 22.98g/mol

Molecular mass of Cl = 35.5g/mol

To find the percentage mass of Na in NaCl we divide the molecular mass of Na by the total number of atoms of Na present in the compound which in this case is 1 by the molecular mass of the compound

%Na = molecular mass of Na / molecular mass of compound.

%Na =( 22.98 / 58.44 ) * 100

% Na = 0.3932 * 100 = 39.32%

The mass percent of Na present in NaCl is 39.32%

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In the reaction of nitrogen gas with oxygen gas to produce nitrogen oxide, what is the effect of adding more oxygen gas to the i
Alborosie
<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

The reaction is;

N₂(g) + O₂(g) ⇆ 2NO(g)

  • When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
  • From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
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3 0
3 years ago
15.1 L N2 at 25 °C and 125 kPa and 44.3 L O2 at 25 °C and 125 kPa were transferred to a tank with a volume of 6.25 L. What is th
Y_Kistochka [10]

Answer:

The total pressure of the mixture in the tank of volume 6.25 litres at 51°C  is 1291.85 kPa.

Explanation:

For N2,

                Pressure(P₁)=125 kPa

                  Volume(V₁)=15·1 L

                Temperature (T₁)=25°C=25+273 K=298 K

Similarly, for Oxygen,

                   Pressure(P₂)= 125 kPa

                   Volume(V₂)= 44.3 L

                  Temperature(T₂)=25°C= 298 K

Then, for the mixture,

              Volumeof the mixture( V)= 6.25 L

                                     Pressure(P)=?

                Temperature (T)= 51°C = 51+273 K=324 K

Then, By Combined gas laws,

                                 \frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}

                      or, \frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}

                     or, 6.34+18.58=\frac{P*6.25}{324}

                     or, P=\frac{24.92*324}{6.25}

                        ∴P=1291.85 kPa

So the total pressure of the mixture in the tank of volume 6.25 litres at 51°C  is 1291.85 kPa.

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SOVA2 [1]

Answer:

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