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3 years ago

One of the most worrisome waste products of a nuclear reactor is plutonium

1 answer:

4 0

One of the most worrisome waste products of a nuclear reactor is plutonium 239 (239Pu). This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus (4He +... Q: One of the most worrisome waste products of a nuclear reactor is plutonium 239 (239Pu).

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Compare the rates of effusion for carbon monoxide and carbon dioxide gases

Airida [17]

According to Grahams law the rate of effusion of a gas is inversely proportional to the square root of it's molecular weight. The rate of diffusion is the measure of rate at which two gases mix. From this law we can say that for the two gases carbon monoxide and carbon dioxide, the rate of effusion of carbon monoxide is greater than that of carbon dioxide, this is because carbon monoxide is lighter (28 g) compared to carbon dioxide (44 g).

7 0

2 years ago

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and

lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

= 25kg

Mass ratio of oil to solution $Y_{F}$ = $\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg

therefore $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$ = $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75= E₁ × N₁

75 = E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0

2 years ago

I need you to solve for me plzzzzzz

Sergeu [11.5K]

Answer:

.0556 L

Explanation:

First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.

Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).

Finally, the answer will be .0556 L.

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3 0

2 years ago

Copper(ii) chloride reacts with sodium nitrate to produce copper(ii) nitrate and sodium chloride.

zlopas [31]

Answer:

CuCl2 + 2NaNO3 - > Cu(NO3)2 + 2NaCl

5 0

1 year ago

A poundal is the force required to accelerate a mass of 1 lbm at a rate of 1 ft/s2, and a slug is the mass of an object that wil

Darya [45]

Remember the formula as per the second Law of Newton: F = m*a

And also remember that the weight is the force with which the mass is attracted by the planet (or satellite in the case of the moon).

With that information you can answer the questions:

a) Weight = F = m*a

m = 175 slugs = 175 lbm

i) Earth

a = 32.17 ft/s^2

Weight on Earth = 175 lbm * 32.17 ft / s^2 = 5,629.75 poundal

ii) Moon

a = [1/6] 32.17 ft/s^2

Weight on the Moon = [1/6]*5,629.75 poundal = 938.29 poundal

b) Force = 355 poundal
m = 25.0 slug

a in m/s^2 = ?

First calculate the force in ft/s^2

F = m*a => a = F/m = 355 poundal / 25.0 slug = 14.2 ft/s^2

Conversion:

14.2 ft / s^2 * [ 0.3048 m/ft] = 4.32816 m/s^2

Answer: 4.33 m/s^2

8 0

2 years ago