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3 years ago

If you start with 100 grams of a radionuclide, how much will you have after two half-lives have passed?

Chemistry

2 answers:

Contact [7]3 years ago

8 0

After first half-life you will have 100/2=50g. After second half-life you will have 50/2=25g.

zubka84 [21]3 years ago

6 0

25 grams.

A half life is the amount of time it takes for an isotope to decay half of itself.

So after one half life, there will be half of the original amount.

100/2=50

After another half life, there will be half of that amount.

50/2 = 25

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Egg white, marshmallow, and smoke from a smokestack are examples of _____. suspensions colloids solutions

d1i1m1o1n [39]

They are an example of colloids .....

6 0

3 years ago

Please calculate the mixed volume (Vmix) when 0.300 mol of ethanol and 0.200 mol of water are mixed. Hint: The partial molar vol

Grace [21]

Answer:

If we assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively, Vmix = 20.5 cm³.

Explanation:

The molar volume of a substance is the ratio between the volume and the number of moles of the substance. It represents the volume that 1 mol of it occupies. Because we don't have access to page 24, let's assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively.

The volume of mixture (Vmix) is the sum of the volume of each substance, which is the number of moles multiplied by molar volume, so:

Vmix = 0.300*57 + 0.200*17

Vmix = 17.1 + 3.4

Vmix = 20.5 cm³

7 0

3 years ago

If 5.57 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 28 ∘C, and the tube is heated to 310 ∘C, the Ag2

Snezhnost [94]

Answer: Total pressure = 7293.2 torr or 9.60 atm

Explanation:

<em>Total pressure = partial pressure of nitrogen + partial pressure of oxygen</em>

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: <em>P₁/T₁=P₂/T₂</em>

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O ----> 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

7 0

3 years ago

What is the limiting factor in determining the accumulation of siliceous ooze/calcareous ooze, respectively?

zhannawk [14.2K]

Answer:

productivity and water depth

Explanation:

The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.

The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.

Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.

8 0

3 years ago

Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)

inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0

3 years ago