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2. Which one of the following chemical equations is balanced? A. 2NaCl + H2SO4 → HCL + NaSO4 B. NH3 + H2O → 2NH4OH C. 2Na + S →

Lady_Fox [76]

D. KOH + H₂SO₄ → KHSO₄ + H₂O

6 0

2 years ago

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

2 years ago

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.

azamat

Answer:

(a) $m_{gold}=7.322g$

(b)

$V_{gold}=0.379cm^3$

$V_{copper}=0.122cm^3$

(c) $\rho _{coin}=15.94g/cm^3$

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

$m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g$

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

$V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3$

$m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3} \\\\V_{copper}=0.122cm^3$

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

$\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\ \\\rho _{coin}=15.94g/cm^3$

Best regards.

3 0

3 years ago

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Describe a group of family on the periodic table

lana66690 [7]

Metals are all in the center of the periodic table and are all made up of metals

6 0

3 years ago

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A student combines 20.0 grams of hydrogen and 20.0 grams of oxygen in a reaction. According to the law of conservation of mass,

jekas [21]

Answer:

A)20.0grams

Explanation:

4 0

2 years ago

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