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3 years ago

A Rock is determined to be 1.426 billion years old. How much uranium -235 remains in the rock?

Chemistry

2 answers:

beks73 [17]3 years ago

4 0

3.5930 is the answer to the question
Hope this helped

Alexxandr [17]3 years ago

3 0

Answer:3.5930 million

Explanation:

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Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

7 0

4 years ago

The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample changed

eimsori [14]

The answer is A. 6.2 calories

3 0

4 years ago

Read 2 more answers

Is heated water more or less dense than the melted snow And rainwater

san4es73 [151]

Answer:

Heated water is more dense than melted snow because water as liquid is denser than ice. Ice floats on water, which means that it has less density than water. Heated water is warmer and more dense than melted ice.

Explanation:

7 0

3 years ago

How many molecules are in 10 moles of Hydrogen?

vivado [14]

Answer:3.2

Mass:32

Molecular weight: 10

Moles: 3.2

8 0

3 years ago

A mixture containing 0.477 mol he(g), 0.265 mol ne(g), and 0.115 mol ar(g) is confined in a 7.00-l vessel at 25 ∘c. part a calcu

enot [183]

Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm

Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.

total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He - $\frac{0.477}{0.857} = 0.557$
mole fraction of Ne - $\frac{0.265}{0.857} = 0.309$
mole fraction of Ar - $\frac{0.115}{0.857} = 0.134$
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm

6 0

3 years ago