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3 years ago

calculate the number of grams of magnesium in 50g of magnesium sulfide. Use percentage composition and gravimetric factor.

Chemistry

1 answer:

faltersainse [42]3 years ago

3 0

Answer:

50 gram of MgS contains =21.56 gram of Mg

Explanation:

Molecular mass of MgS is 56.37 g/mol.

Atomic mass of 1 mole Mg is 24.305

Therefore 56.37 gram of MgS contains 24.305 gram of Mg

1 gram of MgS contains $\frac{24.305}{56.37}$ gram of Mg

50 gram of MgS contains $\frac{24.305}{56.37}$ ×50 gram of Mg

=21.56 gram of Mg

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What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?

katrin2010 [14]

As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane

Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ

X = 0.970 moles of Ocatne

So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol

Mass = 110.83 g of Octane

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2 years ago

Torrey's neighbor told her that a marble rolling down a hill increases in velocity as it rolls down, but does not increase in ki

ziro4ka [17]

Answer:

Torrey's neighbour is incorrect because increase in kinetic energy is proportional to velocity. If the velocity increases so will the object's kinetic energy. Because the mass is constant, if the velocity increases, so does the kinetic energy.

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2 years ago

Which of the following are second-order reactions? a. C2H6 → 2 CH3 b. 2 N2O5 → 4 NO2 + O2 c. 2 HI → H2 + I2 d. 2 N2O → 2 N2 + O2

Klio2033 [76]

Answer:

The answer is b, c, d, e

Explanation:

b. 2 N2O5 → 4 NO2 + O2

r = k [N2O5]^2 --> Second-order regarding global reaction

c. 2 HI → H2 + I2

r = k [HI]^2 --> Second-order regarding global reaction

d. 2 N2O → 2 N2 + O2

r = k [N2O]^2 --> Second-order regarding global reaction

e. 2 NO2 → 2 NO + O2

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2 years ago

Please help ASAP! I’m lost right now.

SCORPION-xisa [38]

1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1

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3 years ago

What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M

Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0

2 years ago