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3 years ago

What is a problem commonly associated with nuclear power facilities?

1 answer:

5 0

(1) False, lots of energy is actually produced from nuclear fuel, if we didn't get much then we probably wouldn't use it
(2) False, its burning coal that contributes to acid rain, since it contains sulfur
(3) False again, we can control the reaction with aptly named control rods, which are typically made of boron, to absorb some of the neutrons flying around in the chain reaction
(4) True, radioactive waste is very difficult to dispose of, and is also very dangerous. Sources of radiation can remain so for millions of years

You might be interested in

What is shown in the diagram below?

Colt1911 [192]

B I think is right. Hope this helps!

3 0

3 years ago

Read 2 more answers

Osmotic pressure Π is given by the relation:Π = iMRTwhere i is the van’t Hoff factor, M is the concentration of solute, R is the

lions [1.4K]

Answer: The concentration of solute is 0.503 mol/L

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=icRT$

where,

$\pi$ = osmotic pressure of the solution = 24 atm

i = Van't hoff factor = 2 (for NaCl)

c = concentration of solute = ?

R = Gas constant = $0.08\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$24atm=2\times c\times 0.08\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\c=0.503mol/L$

Hence, the concentration of solute is 0.503 mol/L

5 0

3 years ago

I don’t understand how to solve this and I don’t get ionic equation

Elden [556K]

Explanation:

Spectator ions refers to elements/molecules that do not form part of the reaction. Basically, these elements/molecules are in the same state in the reactant and product side. From the given reaction, Ca and SO4 change their state from aqueous to solid and Na and Cl do not change their state. Therefore, Na and Cl are spectator ions and Ca and SO4 are not spectator ions.

Answer:

The correct option is C.

8 0

1 year ago

How many grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C?

jolli1 [7]

Answer:

22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law:

P * V = n * R * T

where R is the molar constant of the gases and n the number of moles.

In this case you know:

P= 7.5 atm
V= 2.3 L
n= ?

R= 0.082 $\frac{atm*L}{mol*K}$

T= 24 °C= 297 °K (being 0°C=273°K)

Replacing:

$7.5 atm* 2.3 L=n*0.082 \frac{atm*L}{mol*K} *297K$

Solving:

$n=\frac{7.5 atm* 2.3 L}{0.082 \frac{atm*L}{mol*K} *297K}$

n=0.708 moles

Knowing that oxygen gas is a diatomic gas of molecular form O₂ and its mass is 32 g / mole, you can apply the following rule of three: if 1 mole contains 32 grams, 0.708 moles, how much mass will it have?

$mass=\frac{0.708 moles*32 grams}{1mole}$

mass= 22.656 grams

22.656 grams of oxygen gas are there in a 2.3L tank at 7.5 atm and 24° C

8 0

3 years ago

Read 2 more answers

50.0 mL solution of 0.160 M potassium alaninate ( H 2 NC 2 H 5 CO 2 K ) is titrated with 0.160 M HCl . The p K a values for the

scZoUnD [109]

Answer:

a) 6.12

b) 1.87

Explanation:

At the onset of the equivalence point (i.e the first equivalence point); alaninate is being converted to alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ ------> $H_3}^+NC_2H_5CO^-_2$

1 mole of alaninate react with 1 mole of acid to give 1 mole of alanine;

therefore 50.0 mL of 0.160 M alaninate required 50.0 mL of 0.160M HCl to reach the first equivalence point.

The concentration of alanine can be gotten via the following process as shown below;

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{initial moles of alaninate}{total volume}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+50.0mL)}$

$[H_3}^+NC_2H_5CO^-_2]$ = $\frac{8}{100mL}$

$[H_3}^+NC_2H_5CO^-_2]$ = 0.08 M

Alanine serves as an intermediary form, however the concentration of $H^+$ and the pH can be determined as follows;

$[H^+]$ = $\sqrt{\frac{K_{a1}K_{a2}{[H_3}^+NC_2H_5CO^-_2]+K_{a1}K_w}{ K_{a1}{[H_3}^+NC_2H_5CO^-_2] } }$

$[H^+]$ = $\sqrt{\frac{ (10^{-pK_{a1})}(10^{-pK_{a2})}(0.08)+(10^{-pK_{a1})}(1.0*10^{-14})} {(10^{-pK_{a1}})+(0.08)} }$

$[H^+]$ = $\sqrt{\frac{ (10^{-2.344})(10^{-9.868})(0.08)+(10^{-2.344})(1.0*10^{-14})} {(10^{-2.344})+(0.08)} }$

$[H^+]$ = $7.63*10^{-7}M$

pH = - log $[H^+]$

pH = $-log[7.63*10^{-7}]$

pH= 6.12

Therefore, the pH of the first equivalent point = 6.12

b) At the second equivalence point; all alaninate is converted into protonated alanine.

$H_2NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO^-_2$

$H^+_3NC_2H_5CO^-_2$ $+$ $H^+$ -----> $H^+_3NC_2H_5CO_2H$

Here; we have a situation where 1 mole of alaninate react with 2 moles of acid to give 1 mole of protonated alanine;

Moreover, 50.0 mL of 0.160 M alaninate is needed to produce 100.0mL of 0.160 M HCl in order to achieve the second equivalence point.

Thus, the concentration of protonated alanine can be determined as:

$[H^+_3NC_2H_5CO_2H]$ = $\frac{initial moles of alaninate}{total volume}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{(50.0mL)*(0.160M)}{(50.0mL+100.0mL)}$

$[H^+_3NC_2H_5CO_2H]$ = $\frac{8}{150}$

$[H^+_3NC_2H_5CO_2H]$ = 0.053 M

The pH at the second equivalence point can be calculated via the dissociation of protonated alanine at equilibrium which is represented as:

$H^+_3NC_2H_5CO_2H$ ⇄ $H^+_3NC_2H_5CO^-_2$ $+$ $H^+$

(0.053 - x) x x

$K_{a1}$ = $\frac{[H^+] [H^+_3NC_2H_5CO^-_2]}{[H^+_3NC_2H_5CO_2H]}$

$10^{-PK_{a1}}$ = $\frac{x*x}{(0.053-x)}$

$10^{-2.344} =\frac{x^2}{(0.053-x)}$

$0.00453 = \frac{x^2}{(0.053-x)}$

$0.00453(0.053-x) =x^2$

$x^2+0.00453x-(2.4009*10^{-4})$

Using quadratic equation formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

we have:

$\frac{-0.00453+\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$ OR $\frac{-0.00453-\sqrt{(0.00453)^2-4(1)(-2.4009*10^{-4})} }{2(1)}$

= 0.0134 OR -0.0179

So; we go by the positive integer which says

x = 0.0134

So $[H^+]=[H_3^+NC_2H_5CO^-_2]= 0.0134 M$

pH = $-log[H^+]$

pH = $-log[0.0134]$

pH = 1.87

Thus, the pH of the second equivalent point = 1.87

3 0

4 years ago