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3 years ago

Whats the molar mass of CO

Chemistry

2 answers:

g100num [7]3 years ago

7 0

28.01 g/mol

hope that helped

tamaranim1 [39]3 years ago

5 0

Answer:

28.01

Explanation:

this is the molar mass

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The formula of nitrogen oxide is NO, of nitrogen dioxide is NO_2. Write a balanced equation for the reaction of nitrogen oxide w

denis23 [38]

Answer:

a) 0,5 mol O₂; 1 mol NO₂

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) NO = 30g

O₂ = 16g

NO₂ = 46g

d) 7,41g HCl

e) 107,7 g/mol

f) 0,0312 moles of O₂; 0,999g of O₂; 699 mL at STP or 803 mL

ii. 51%

g) 32,6L

Explanation:

a) For the reaction:

2 NO + O₂ → 2 NO₂

For 1 mole of NO there are consumed:

1 mol NO × $\frac{1molO_{2}}{2molNO}$ = 0,5 mol of O₂

And produced:

1 mol NO × $\frac{2molNO_{2}}{2molNO}$ = 1 mol of NO₂

b) By ideal gas law:

V = nRT/P

Where n is moles of each compound; R is gas constant (0,082atmL/molK); T is temperature (273,15 K at state conditions); P is pressure (1 atm at STP) and V is volume in liters. Replacing each moles for each compound:

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) The mass of each compound are:

1 mol NO× $\frac{30 g}{1molNO}$ = 30g

0,5 mol O₂× $\frac{32 g}{1molO_{2}}$ = 16g

1 mol NO₂× $\frac{46 g}{1molNO_{2}}$ = 46g

d) Using:

n = PV / RT

Moles of 4,55 L of HCl (using the values of P = 1 atm; R = 0,082atmL/molK; T = 273,15K) are:

0,203 moles of HCl. In grams:

0,203 mol HCl× $\frac{36,46 g}{1molHCl}$ = 7,41 g of HCl

e) Using:

δRT/P = MW

Where δ is density in g/L (4,81 g/L); R is gas constant (0,082atmL/molK); T is temperature (273,15K); P is pressure (1 atm)

And MW is molecular mass: 107,7 g/mol

f) For the reaction:

2 KClO₃ → 2 KCl + 3 O₂

2,550 g of KClO₃ are:

2,550 g of KClO₃× $\frac{1mol}{122,55 gKClO_{3}}$ = 0,0208 moles of KClO₃

When these moles reacts completely produce:

0,0208 moles of KClO₃× $\frac{3 mol O_{2}}{2 molKClO_{3}}$ = 0,0312 moles of O₂

In grams:

0,0312 moles of O₂ × $\frac{32g}{1 molO_{2}}$ = 0,999g of O₂

V = nRT/P

At STP, n = 0,0312 mol; R = 0,082atmL/molK;T= 273,15K; P = 1atm; V = 0,699L ≡ 699mL

At 29 °C (302,15K) and 732 torr (0,963 atm)

V = 0,803L ≡ 803mL

ii. 182 mL ≡ 0,182L of O₂ are:

n = PV/RT

moles of O₂ are 7,07x10⁻³. Moles of KClO₃ are:

7,07x10⁻³ moles of O₂× $\frac{2 mol KClO_{3}}{3 molO_{2}}$ = 0,0106 mol KClO₃. In grams:

0,0106 moles of KClO₃× $\frac{122,55 g}{1molKClO_{3}}$ = 1,300 g of KClO₃.

Thus, percent by mass of KClO₃ in the mixture is:

1,300g/2,550g ×100 = 51%

g. Combined gas law says that:

$\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}$

Where:

P₁ = 755 torr; V₁ = 35,9L; T₁ = 26°C (299,15 K); P₂ = 760 torr (STP): T₂ = 273,15K (STP) V₂ = 32,6 L

I hope it helps!

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