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Tema [17]
4 years ago
13

A geological sample is found to have a Pb-206/U-238 mass ratio of 0.337/1.00. Assuming there was no Pb-206 present when the samp

le was formed, how old is it? The half-life of U-238 is 4.5 × 109 years.
Chemistry
2 answers:
DENIUS [597]4 years ago
5 0

Answer:

7.1 x 10^9 years

Explanation:

This is the answer according to Mastering Chemistry.

azamat4 years ago
4 0

Answer:

Age=2.52*10^9 years

Explanation:

To determine the age, we need to know how much U have become Pb, so first we have to calculate the Pb moles present in a sample:

n_{Pb}=\frac{0.337g}{206g/mol}=0.00164 mol

n_{U}=\frac{1 g}{238g/mol}=0.0042 mol

The percentage of U degradation:

P=\frac{n_{Pb}}{n_{Pb}+n_{U}}

P=\frac{0.00164}{0.00164+0.0042}=0.28

Assuming that the life time is linear:

Age=\frac{4.5*10^9 years}{0.5 life time}*0.28 life time

Age=2.52*10^9 years

You might be interested in
Through electrolysis, a student collects 6.6 g of hydrogen gas and 52.4 g of oxygen gas. The reaction was allowed to proceed to
Kaylis [27]

Answer: 59 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2H_2O\rightarrow 2H_2+O_2

Given: mass of hydrogen = 6.6 g

mass of oxygen = 52.4 g

Mass of products = Mass of hydrogen + mass of oxygen = 6.6 +52.4 = 59 g grams

Thus mass or reactant = mass of water

Mass of reactants = mass of products = 59 g

Thus the mass of water initially present was 59 g.

3 0
4 years ago
hydrogen decomposes quickly when exposed to light. the reaction for decomposing hydrogen peroxide is:
Svet_ta [14]

The decomposition reaction for hydrogen peroxide is given below:

                                     2H_{2} O_{2} → 2H_{2} O + O_{2}

This is a decomposition reaction.

Reactions in which a reactant breaks into two or more products are known as Decomposition reactions.

                                          AB → A + B

here, AB represents the reactant that begins the reaction, and A and B represent the products of the reaction

The decomposition reaction of decomposing hydrogen peroxide is exothermic. When the hydrogen peroxide undergoes a decomposition reaction, heat is also released along with water and oxygen.

Hence the  reaction for decomposing hydrogen peroxide is :

         2H_{2} O_{2} → 2H_{2} O + O_{2}

If you need to learn more about Decomposition reaction, click here

brainly.com/question/16987748?

#SPJ4

5 0
1 year ago
Name a device or other object whose major purpose is to convert
Alenkinab [10]

Answer:

Storage battery oe cell

Explanation:

The storage battery or cell is a device that coverts chemical energy into light energy when required.

During charging of battery, there is some chemical changes in the battery and absorb energy. The absorbed energy is converted into electrical energy when connected to an external load.

Hence, the correct answer is "Storage battery or cell".

4 0
3 years ago
At 333 k, which of the pairs of gases below would have the most nearly identical rates of effusion?
rjkz [21]
E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
a) M(N</span>₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, <span>CO and N</span>₂ <span>have iexact molecular masses and will effuse at nearly identical rates.</span>
8 0
3 years ago
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
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