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3 years ago

If freshwater is added to a region, what happens to seawater density and salinity?

Chemistry

1 answer:

azamat3 years ago

7 0

Answer:

The density and salinity goes down.

Explanation:

Salt water is denser than pure water. As fresh water is added to salt water the concentration of salt goes down. Since salinity is the amount of salt in the water that would go down with the concentration. This would also mean that the density goes down since there is less salt per unit of volume which makes it so that there is less mass per unit of volume. The formula for density is d=m/v so as the mass goes down per unit of volume so does the density.

Let me know if anything is unclear in the comments and we can try to work through it.

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If a sample has 25% parent and 75% daughter, how many halflifes have gone by?

Anit [1.1K]

Answer:

Two half lives.

Explanation:

It is known that the decay of isotopes and radioactive material obeys first order kinetics.

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

That means for a sample 100% to decay to 50 % it will take one half-life, and to decay the remaining 50% to 25% it will take another half-life.

So, for a sample has 25% parent and 75% daughter it will have two half-lives.

6 0

3 years ago

Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette sm

vladimir2022 [97]

Answer:

pH = 9.29

Explanation:

C5H5N + H2O ↔ C5H5NH+ + OH-

∴ pKb = 8.75 = - Log Kb

⇒ Kb = 1.7783 E-9 = [OH-][C5H5NH+]/[C5H5N]

mass balance:

⇒ <em>C</em> C5H5N = 0.215 M = [ C5H5N ] + [C5H5NH+]

charge balance:

⇒ [OH-] = [C5H5NH+]

⇒ Kb = 1.7783 E-9 = [OH-]² / ( 0.215 - [OH-] )

⇒ [ OH-]² + 1.7783 E-9[OH-] - 3.8233 E-10 = 0

⇒ [OH-] = 1.9552 E-5 M

⇒ pOH = 4.708

∴ 14 = pH + pOH

⇒ pH = 9.29

5 0

3 years ago

Assume the volume of the crew cabin is 74,000 L. The pressure is maintained at around 1.00 atm, ideally with an 80% nitrogen and

hjlf

Answer:

The number of moles of oxygen present in the crew cabin at any given time is 615.309 moles

Explanation:

The given parameters are;

Volume of the crew cabin = 74,000 L

Pressure of the crew cabin = 1.00 atm

Percentage of nitrogen in the mixture of gases in the cabin = 80%

Percentage of oxygen in the mixture of gases in the cabin = 20%

Temperature of the cabin = 20°C = 293.15 K

Therefore, volume of oxygen in the crew cabin = 20% of 74,000 L

Hence, volume of oxygen in the crew cabin = $\frac{20}{100} \times 74,000 \, L = 14,800 \, L$

From the universal gas equation, we have;

$n = \frac{P \times V}{R \times T}$

Where:

n = Number of moles of oxygen

P = Pressure = 1.00 atm

V = Volume of oxygen = 14,800 L

T = Temperature = 293.15 K

R = Universal Gas Constant = 0.08205 L·atm/(mol·K)

Plugging in the values, we have;

$n = \frac{1 \times 14,800 }{0.08205 \times 293.15 } = 615.309 \, moles$

The number of moles of oxygen present in the crew cabin at any given time = 615.309 moles.

3 0

4 years ago

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Anna [14]

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5 0

3 years ago

What quantum numbers specify a 5p orbital?

mezya [45]

n = 5

l = 0,1,2,3,4

ml = -4,-3,-2,-1,0,1,2,3,4,

ms = +1/2 and -1/2

8 0

3 years ago