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Goryan [66]
3 years ago
15

What is the half life of the element in the picture HELP BRAINLIEST

Chemistry
1 answer:
stira [4]3 years ago
5 0

Answer:

6 days

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 100 mg

Amount remaining (N) = 6. 25 mg

Time (t) = 24 days

Half life (t½) =?

Next, we shall determine the decay constant. This can be obtained as follow:

Original amount (N₀) = 100 mg

Amount remaining (N) = 6. 25 mg

Time (t) = 24 days

Decay constant (K) =?

Log (N₀/N) = kt / 2.303

Log (100/6.25) = k × 24 / 2.303

Log 16 = k × 24 / 2.303

1.2041 = k × 24 / 2.303

Cross multiply

k × 24 = 1.2041 × 2.303

Divide both side by 24

K = (1.2041 × 2.303) / 24

K = 0.1155 /day

Finally, we shall determine the half-life of the isotope as follow:

Decay constant (K) = 0.1155 /day

Half life (t½) =?

t½ = 0.693 / K

t½ = 0.693 / 0.1155

t½ = 6 days

Therefore, the half-life of the isotope is 6 days

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Many of the lunar craters are due to volcanic eruptions true or false
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150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac
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Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

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