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2 years ago

A sample of 1.55 g of iron ore is dissolved in an acid solution in which the iron is converted into

Chemistry

1 answer:

solmaris [256]2 years ago

4 0

Answer:a) 8H2SO4 + 2KMnO4 + 10FeSO4 → 5Fe2(SO4)3 + 8H2O + 2MnSO4 + K2SO4

B) nKMno4 = 0.001859 mol

=>nFeso4=0.009295 mol

nFe= 0.009295 mol

mFe=0.52052 g

=>percentage of iron in the sample: 33.5819%

Explanation:a) 8H2SO4 + 2KMnO4 + 10FeSO4 → 5Fe2(SO4)3 + 8H2O + 2MnSO4 + K2SO4

B) nKMno4 = 0.001859 mol

=>nFeso4=0.009295 mol

nFe= 0.009295 mol

mFe=0.52052 g

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A soft drink contains 11.5% sucrose (C12H22O11) by mass. How much sucrose, in grams, is contained in 355 mL (12 oz) of the soft

luda_lava [24]

Answer:

42.5 g

Explanation:

Calculate the mass of the soft drink given the density and volume:

355 mL × 1.04 g/mL = 369.2 g

Now calculate the mass of sucrose given the percentage:

0.115 × 369.2 g = 42.46 g

Rounded to 3 significant figures, the mass is 42.5 g.

5 0

2 years ago

Read 2 more answers

Alveolar air (a mixture of nitrogen, oxygen, and carbon dioxide) has a total pressure of 0.998 atm. If the partial pressure of o

inn [45]

Answer:

The partial pressure of carbon dioxide is 22.8 mmHg

Explanation:

Dalton's Law is a gas law that relates the partial pressures of the gases in a mixture. This law says that the pressure of a gas mixture is equal to the sum of the partial pressures of all the gases present.

In this case:

Ptotal=Pnitrogen + Poxygen + Pcarbondioxide

You know that:

Ptotal= 0.998 atm
Pnitrogen= 0.770 atm
Poxygen= 0.198 atm
Pcarbondioxide= ?

Replacing:

0.998 atm=0.770 atm + 0.198 atm + Pcarbondioxide

Solving:

Pcarbondioxide= 0.998 atm - 0.770 atm - 0.198 atm

Pcarbondioxide= 0.03 atm

Now you apply the following rule of three: if 1 atm equals 760 mmHg, 0.03 atm how many mmHg equals?

$Pcarbondioxide=\frac{0.03 atm*760 mmHg}{1 atm}$

Pcarbondioxide= 22.8 mmHg

The partial pressure of carbon dioxide is 22.8 mmHg

6 0

2 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago

A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the s

yKpoI14uk [10]

Answer:

27.0

Explanation:

Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.

7 0

2 years ago

Given that 25.0 mL of mercury has a mass of 340.0 g, calculate (a) the density of mercury and (b) the mass of 120.0 mL of mercur

natita [175]

Answer :

(a) The density of mercury is, 13.6 g/ml

(b) The mass of 120.0 ml of mercury is, 1632 grams

Explanation :

(a) Now we have to calculate the density of mercury.

Given :

Volume of mercury = 25.0 ml

Mass of mercury = 340.0 g

Formula used :

$\text{Density of mercury}=\frac{\text{Mass of mercury}}{\text{Volume of mercury}}$

$\text{Density of mercury}=\frac{340.0g}{25.0ml}=13.6g/ml$

Therefore, the density of mercury is, 13.6 g/ml

(b) Now we have to calculate the mass of 120.0 ml of mercury.

As, 25.0 ml of mercury has mass = 340.0 g

So, 120.0 ml of mercury has mass = $\frac{120.0ml}{25.0ml}\times 340.0g=1632g$

Therefore, the mass of 120.0 ml of mercury is, 1632 grams

3 0

2 years ago