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2 years ago

Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.

1 answer:

5 0

The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%

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Chlorine gas was first prepared in 1774 by the oxidation of NaCl with MnO2:

irina1246 [14]

Answer:

0.5

Explanation:

2NaCl(s) + 2H2SO4(l) + MnO2(s) → Na2SO4(s) + MnSO4(s) + 2H2O(g) + Cl2(g)

Using ideal gas equation,

PV = nRT

28.7torr

Converting torr to atm,

= 0.0378atm

V = 0.597L

T = 27 °C

= 300 K

a) PV = nRT

(0.0378atm) * (0.597L) = n(0.0821) * (300k)

= 0.000915 mol

moles of water and chlorine = 0.000915 mol

From the above equation, the ratio of water to chlorine = 1 : 2

Therefore, mole of chlorine = 0.000915/2

= 0.000458 mol

mole fraction = moles of specie/moles of all the species present

= 0.000458/0.000915

= 0.5

5 0

2 years ago

6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13

vodomira [7]

Answer:

The molecular formula of X is given as $C_7 H_6 O_3$

Explanation:

$Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$$

$mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

$moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

$\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$$

So the empirical formula is $C_7 H_6 O_3$

Empirical formula mass

$=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g$

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

$=1 \times C_7 H_6 O_3$

Compound X = $C_7 H_6 O_3$ is the Answer

8 0

2 years ago

Two containers hold the same radioactive isotope. Container A contains 1000 atoms, and container B contains 500 atoms. which con

enyata [817]

Answer:

The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.

Explanation:

From the question,

Container A contains 1000 atoms

Container B contains 500 atoms

The rate of decay of atoms in container A is greater than the rate of decay of atoms in container B.

The reason for such is due to the difference in the concentration of the isotopes. Container A which contains higher number of atoms will have the more changes of the release of the neutron as the changes of the hitting and splitting increases as the density of the atoms increases.

Thus, the atoms in the container A will therefore decay faster than the atoms in the container B.

8 0

2 years ago

SbF5 mass ratio and atomic ratio

Marina CMI [18]

Answer:

SbF5

Mass ratio = 1.28 : 1

And atomic ratio = 1 : 5

7 0

2 years ago

A load of bauxite has a density of 3.11 . If the mass of the load is 220. metric tons, how many dump trucks, each with a capacit