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A chemist wishing to do an experiment requiring 47Ca2+(half-life = 4.536 days) needs 5.00 g of the nuclide. What mass of 47CaCO

Artyom0805 [142]

Answer: The answer is 6.78 grams.

Explanation: The equation used for solving this type of problems is:

$\frac{N}{N_0}=(\frac{1}{2})^n$

where, $N_0$ is the initial amount of radioactive substance, N is the remaining amount and n is the number of half lives.

Number of half lives is calculated on dividing the given time by the half life.

n = time/half life

Time is given as 48.0 hours and the half life is given as 4.536 days. let's make the units same and for this let's convert the half life from days to hours.

$4.536days(\frac{24hours}{1day})$

= 108.864 hours

So, $n=\frac{48.0}{108.864}$ = 0.441

Since 5.00 g is the required amount when the radioactive substance is delivered to the scientist, it would be the final amount that is N. We need to calculate the initial amount. Let's plug in the values in the equation:

$\frac{5.00}{N_0}=(\frac{1}{2})^0^.^4^4^1$

$\frac{5.00g}{N_0}=0.737$

$N_0=\frac{5.00g}{0.737}$

$N_0$ = 6.78 g

So, 6.78 g of the radioactive substance needs to be ordered.

4 0

4 years ago

The hydrogen halide gases all have the same general formula, HX, where X can be a halogen (F, CI, Br, or I). HCl 2 effuses times

Natali [406]

Answer:

Explanation:

We collected the gas over water; to obtain the true pressure of the gas, we have to correct for the vapor pressure of water. We look up the vapor pressure of water at 24 ºC: 24 torr. The pressure of the evolved gas is, therefore, 738 −24 = 714 torr. 2. We want the molar mass of the

8 0

3 years ago

BaS + PtF2 = BaF2 + PtS

dedylja [7]

I think it is
Barium Sulfide + Platinum Difluoride = Barium Fluoride + Platinum Sulfide

3 0

4 years ago

4. Complete the following equations:

lara31 [8.8K]

Answer:

2NaCl + CuCO3

FeCl2 + BaSO4

CuCO3 + Ca(NO3)2

Explanation:

Presumably this is a double replacement reaction.

A+B + C+D → A+D + C+B

It seems I may be wrong so please try to work out the problem yourself to double check, keeping in mind the charges of each compound.

4 0

3 years ago

A 50. g sample of dry iron(II) sulfate has which mass if converted to the pentahydrate?

IRISSAK [1]

Answer is: mass of iron(II) sulfate pentahydrate (FeSO₄ · 5H₂O) would be 79,62 g.
m(FeSO₄) = 50,0 g.
m(FeSO₄ · 5H₂O) = ?
m(FeSO₄) : M(FeSO₄) = m(FeSO₄ · 5H₂O) : M(FeSO₄ · 5H₂O).
50 g : 151,9 g/mol = m(FeSO₄ · 5H₂O) : 241,9 g/mol.
m(FeSO₄ · 5H₂O) = 50 g · 241,9 g/mol ÷ 151,9 g/mol.
m(FeSO₄ · 5H₂O) = 79,62 g.
M - molar mass.

8 0

4 years ago

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