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2 years ago

As of January 2009, the USA has produced 60,000 metric tons of nuclear waste in 60 yesrs of operating 104 nuclear power plants.

Chemistry

2 answers:

goblinko [34]2 years ago

5 0

Answer: a. 0.8 tons p. month.

Explanation:

Given: Total nuclear waste = 60,000 metric tons

Time take = 60 years = 12 x 60 months [ 1 year = 12 months]

= 720 months

Total nuclear power plants = 104

Now , Average waste produced by each plant = $\dfrac{\text{Total waste}}{\text{Time x Number of plants}}$

$=\dfrac{60000}{720\times104}\approx0.8\text{ tons per month.}$

Hence, 0.8 tons p. month is produced by each plant.

So, option a. is correct.

Ipatiy [6.2K]2 years ago

3 0

Answer:

A on edge 2021

Explanation:

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Water has a heat of fusion of about 300J/gram and a heat capacity for liquid water of a about 4J/gC. How much heat is needed to

amm1812

Answer:

The total heat required is 3.4 kJ

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. So, the amount of heat a body receives or transmits is determined by:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case you know;

Q= ?

c= 4 $\frac{J}{g*C}$

m= 10 g
ΔT= Tfinal - Tinitial= 10 C - 0 C= 10 C

Replacing:

$Q= 4\frac{J}{g*C} *10 g * 10 C$

Solving:

Q1= 400 J

On the other hand, you must determine the heat required to convert 0 ∘ C of ice to 0 ∘ C of liquid water by:

Q2=m*heat of fusion

Q2=10 g* 300 $\frac{J}{g}$

Q2= 3,000 J

The total heat required is:

Q= Q1 + Q2= 400 J + 3,000 J

Q= 3,400 J= 3.4 kJ (1 kJ= 1,000 J)

The total heat required is 3.4 kJ

5 0

3 years ago

) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H

Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

$pH = -log([H_{3}O^{+}])$

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

Trimethyl ammonium:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺

1.0 - x x x

$Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}$

$10^{-pKa} = \frac{x*x}{1.0 - x}$

$10^{-9.80}(1.0 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 0.097 = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01$

Phenol:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x x x

$Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}$

$10^{-10} = \frac{x^{2}}{1.0 - x}$

$1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0$

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00$

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

$\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95$

Therefore, the difference in pH values is 4.95.

I hope it helps you!

7 0

3 years ago

Which diagram shows a pair of electrons that have opposite spins? 011 1 011 o 11h

Neporo4naja [7]

Answer:

1 one is the right........

5 0

2 years ago

Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0

2 years ago

Question 38 (1 point)

Fiesta28 [93]

Answer: Protons because they have a positive charge.

Explanation:

8 0

3 years ago