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Alchen [17]
4 years ago
6

04-10 A power plant burns 100 metric tons of coal per day with a sulfur content of 5%. The coal contains 3% ash (noncombustible

portion). Samples of the ash show that it has a sulfur content of 10% sulfur by weight. Sulfur that is not in the ash is in the flue gases. How many metric tons of SO2 are emitted from the smoke stack each day? Answer: 9.4 ton SO2/d
Chemistry
1 answer:
slava [35]4 years ago
7 0

Answer:

9.4 metric ton of SO₂ are emitted from yhe smoke stack each year.

Explanation:

Total amount = 100 metric ton

Sulphur in the total amount = 5% = 5 metric ton

Ash = 3% = 3 metric ton

Sulphur in the ash = 10% of 3 metric ton = 0.3 metric ton

Sulphur in the flue gases = 5 metric ton - 0.3 metric ton = 4.7 metric ton

Combustion of sulphur: S (s) + O₂ (g) → SO₂ (g)

32 g S _________ 64 g SO₂

4.7 x 10⁶ g ______    x

x = 9.4 x 10⁶ g = 9.4 metric ton of SO₂

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Answer:

B

Explanation:

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A 25.0 mL sample of a solution of a monoprotic acid is titrated with a 0.115 M NaOH solution. The end point was obtained at abou
kati45 [8]

Answer:

The concentration of the acid is about 0.114 M (option E)

Explanation:

Step 1: Data given

Volume of the monoprotic acid = 25.0 mL = 0.025 L

Molarity of the monoprotic acid = ?

Molarity of the NaOH solution = 0.115 M

Volume NaOH = 24.8 mL = 0.0248 L

Step 2: Calculate the concentration

a*Cb * Vb = b * Ca * Va

⇒ a = the coeficient of NaOH = 1

⇒ Cb = the molarity of the acid = TO BE DETERMINED

⇒ Vb = the volume of the acid = 0.025 L

⇒ b = the coefficient of the acid = monoprotic = 1

⇒ Ca = the moalrity of NaOH = 0.115 M

⇒ Va = the volume of NaOH = 0.0248 L

1 * Cb * 0.025 = 1 * 0.115 * 0.0248

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5 0
4 years ago
. The Henry's law constant for helium gas in water at 30 ∘C is 3.7×10−4M/atm; the constant for N2 at 30 ∘C is 6.0×10−4M/atm. a.
ludmilkaskok [199]

Explanation:

1) Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

M_{g}=K_H\times p_{g}

where,

K_H = Henry's constant

p_{He} = partial pressure of gas

a)  K_H=3.7\times 10^{-4} M/atm

p_{He}=2.1 atm

Putting values in above equation, we get:

M_{He}=3.7\times 10^{-4} M/atm\times 2.1 atm = 7.77\times 10^{-4} M

The solubility of helium gas is 7.77\times 10^{-4} M

b)  K_H=6.0\times 10^{-4} M/atm

p_{N_2}=2.1 atm

Putting values in above equation, we get:

M_{He}=6.0\times 10^{-4} M/atm\times 2.1 atm = 1.26\times 10^{-3} M

The solubility of nitrogen gas is 1.26\times 10^{-3} M

2)

a) Mass of solute or methanol , m= 14.7 g

Mass of solvent or water , m'= 186 g

Mass of the solution = M = m + m' = 14.7 g + 186 g = 200.7 g

w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

=\frac{14.7 g}{200.7 g}\times 100=7.32\%

The mass percent of methanol is 7.32%.

b) Molality is defined as moles of solute per kilograms of solvent.

m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}

Moles of methanol = \frac{14.7 g}{32 g/mol}=0.4594 mol

Mass of solvent = 186 g = 0.186 kg

m=\frac{0.4594 mol}{0.186 kg}=2.4610 mol/kg

The molality of methanol is 2.4610 mol/kg.

4 0
3 years ago
An isotope of Cesium -137 has a half-life of 30 years. If 1.0 g of Cesium – 137 disintegrates over a period of 60 years, how man
Nady [450]

Answer:

Explanation:

Radioactive decay follows the equation:

Ln [A] = -kt + ln [A]₀

<em>Where [A] is amount of isotope after time t: Our incognite,</em>

<em>k is rate constant: ln 2 / Half-life = 0.0231 years⁻¹</em>

<em>t are 60 years</em>

<em>[A]₀ is initial amount of isotope: 1.0g</em>

<em />

Replacing:

Ln [A] = -kt + ln [A]₀

Ln [A] = -0.0231 years⁻¹*60 years + ln 1.0g

ln [A] = -1.386

[A] = 0.25g

<em />

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