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tekilochka [14]

2 years ago

11

Carbon exists in nature as three isotopes. Given the atomic masses and relative abundance of the isotopes, what is the average a

tomic mass of carbon? A. 12. 000 amu B. 12. 011 amu C. 13. 002 amu D. 13. 003 amu E. 14. 003 amu.

1 answer:

Shtirlitz [24]2 years ago

3 0

Answer:

B. 12. 011 amu

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Oxygen and hydrogen form the polyatomic hydroxide ion. What is its charge?

prohojiy [21]

Answer:

negative

the chage on hydroxide ions os negative

4 0

3 years ago

Read 2 more answers

Fe2O3 + CO → Fe + CO2 If 2 moles of Fe2O3 react with 9 moles of CO, how many moles of each product are formed?

Y_Kistochka [10]

Answer: B. 4 moles Fe and 3 moles CO2

Explanation:

write a balanced chemical reaction

that is FeO3 + 3 CO → 2 Fe + 3CO2

2 moles of Fe2O3 reacted with 3 x2=6 moles of Co to form Fe and CO2 therefore CO was in excess and Fe2O3 was limiting reagent.

use the mole ratio to determine the moles of each product.

that is the mole ratio 0f Fe2O3 : Fe is 1:2 therefore the moles of Fe = 2x2=4 moles

the mole ratio of Fe2CO3 : CO2 is 1: 3 therefore the moles of Co2 = 2 x3 = 6 moles

8 0

2 years ago

Read 2 more answers

Ammonia and oxygen react to form nitrogen and water.

Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0

3 years ago

A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total intern

Arada [10]

<u>Answer:</u> The volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

<u>Explanation:</u>

To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L$

Converting this value into $m^3$ , we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow (\frac{1m^3}{1000L})\times 154.21L$

$\Rightarrow 0.1542m^3$

To calculate the internal energy, we use the equation:

$U=\frac{3}{2}nRT$

or,

$U=\frac{3}{2}\frac{m}{M}RT$

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.314J/K.mol$

T = temperature = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

6 0

3 years ago

soldier1979 [14.2K]

Answer:

First one is: ammonia

Second one is: calcium hydroxide

Explanation:

4 0

3 years ago