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3 years ago

Calcium-45 has a half-life of 162.7 days. If four half-lives have elapsed, how much time has passed?

2 answers:

7 0

Half-life<span> is the time required for a quantity to reduce to half its initial value. </span><span>If four half-lives have elapsed for calcium-45, then it would be 4x162.7 = 650.8 days have passed. Hope this answers the question. Have a nice day. Feel free to ask more questions.</span>

Len [333]3 years ago

6 0

Answer:

1.78 years

Explanation:

First half life is the time at which the concentration of the reactant reduced to half.

Second half reaction is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/4.

Third half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/8.

Fourth half life is the time at which the remaining concentration reduced to half or the initial concentration reduced to 1/16.

There are four half lives. So, time has passed = 4*162.7 days = 650.8 days = 1.78 years

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Compound X has a molar mass of 416.48 g mol

bazaltina [42]

Answer:

P₂Cl₁₀

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound X = 416.48 g/mol

Percentage of phosphorus (P) = 14.87%

Percentage of Chlorine (Cl) = 85.13%

Molecular formula of X =?

Next, we shall determine the empirical formula of compound X. This can be obtained as follow:

P = 14.87%

Cl = 85.13%

Divide by their molar mass

P = 14.87 / 31 = 0.480

Cl = 85.13 / 35.5 = 2.398

Divide by the smallest

P = 0.480 / 0.480 = 1

Cl = 2.398 / 0.480 = 5

Empirical formula of compound X is PCl₅

Finally, we shall determine the molecular formula of compound X. This can be obtained as follow:

Molar mass of compound X = 416.48 g/mol

Empirical formula = PCl₅

Molecular formula =?

Molecular formula= [Empirical formula]ₙ

[PCl₅]ₙ = 416.48

[31 + (35.5 × 5)]ₙ = 416.48

[31 + 177.5]n = 416.48

208.5n = 416.48

Divide both side by 208.5

n = 416.48 / 208.5

n = 2

Molecular formula = [PCl₅]ₙ

Molecular formula = [PCl₅]₂

Molecular formula = P₂Cl₁₀

Therefore, the molecular formula of compound X is P₂Cl₁₀

4 0

3 years ago

True or false, The atomic number of an element is a whole number that decreases as you read across each row of

torisob [31]

Answer:

I believe it's false because the atomic number is the number of protons in the nucleus of an atom.

5 0

3 years ago

How many moles of hydrogen gas will be produced if 5.00 moles of zinc reactions with an excess amount of sulfuric acid

harkovskaia [24]

Answer:

Moles of Hydrogen produced is 5 moles

Explanation:

The balanced Chemical equation for reaction between zinc and sulfuric acid is :

$Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)$

This equation tells that ; when 1 mole of Zn react with 1 mole of sulfuric acid, it produces 1 mole of zinc sulfate and 1 mole of hydrogen.

Since sulfuric acid is in excess so Zinc is the limiting reagent

(Limiting reagent : Substance which get consumed when the reaction completes, limiting reagent helps in predicting the amount of products formed)

Limiting reagent (Zn) will decide the amount of Hydrogen produced

$Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)$

$1\ mole\ zinc\rightarrow 1\ mole\ H_{2}$

So,

$5\ mole\ zinc\rightarrow 5\ mole\ H_{2}$

Hence moles of Hydrogen produced is 5 moles

3 0

3 years ago

Chloroform (CHCl3), an important solvent, is produced by a reaction between methane and chlorine. CH4(g) + 3 Cl2(g) CHCl3(g) + 3

PIT_PIT [208]

CH₄(g) + 3 Cl₂(g) → CHCl₃(g) + 3 HCl(g)
From the equation we notice that 1 mole of methane produces 1 mole of chloroform:
16 g Methane → 119.38 g Chloroform
? g Methane → 37.5 g Chloroform
by cross multiplication:
= (16 * 37.5) / 119.38 = 5.0 g methane

4 0

3 years ago

Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h

scZoUnD [109]

$44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
CH$

4 0

3 years ago