All categories

3 years ago

After 56 days, a 4.0-g sample of phosphorus-32 contains only 0.25 g of isotope. Based on this data,

Chemistry

2 answers:

Katena32 [7]3 years ago

6 0

Answer:

THE HALF LIFE OF PHOSPHORUS-32 ISOTOPE IS 14 DAYS.

Explanation:

Using the Nerst equation of half life;

t1/2 = t * in(2) / in (No/Nt)

where;

t1/2 = half life = the unknown

t = elapsed time = 56 days

No= initial mass of the sample = 4.0 g

Nt = final mass of the sample = 0.25 g

Substituting the values into the equation, we have;

$t \frac{1}{2} = 56 * \frac{LN2}(LN 4/ 0.25) \\$ t1/2 = 56 * Ln2 / Ln ( 4/0.25)

t1/2 = 56 * 0.6932 / Ln (16)

t1/2 = 56 * 0.6932 / 2.7726

t1/2 = 38.8192 / 2.7726

t1/2 = 14.001 days.

t1/2 = 14 days.

The half life of the phosphorus- 32 would be 14 days

timama [110]3 years ago

5 0

Answer:324 hours (13.5 days)

Explanation:see attached photo

You might be interested in

How many variables are there in science

Sedaia [141]

There are three variables independent, dependent ,and controlled

7 0

3 years ago

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ

NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)

b) The precipitate formed is Ni(OH)₂

c) The limiting reactant is:

$n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles$

$n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles$

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

$n = \frac{2}{1}*0.030 moles = 0.060 moles$

Hence, the limiting reactant is KOH.

d) The mass of the precipitate formed is:

$n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles$

$m = n*M = 0.010 moles*92.72 g/mol = 0.927 g$

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

$C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M$

$C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M$

$C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M$

I hope it helps you!

5 0

4 years ago

Coal, which is primarily carbon, can be converted to natural gas, primarily CH4, by the following exothermic reaction: C(s)+2H2(

LuckyWell [14K]

Answer:

A. The reaction will proceed forward forming more CH4

B. The reaction will proceed forward forming more CH4

C. Since the reaction is exothermic, raising the temperature will cause the reaction to proceed backward, thus forming C and H2.

D. Lowering the volume makes the gas particles to be more close together thereby enhancing their collisions leading to reaction. Therefore the reaction will proceed forward forming more CH4

E. Catalyst only reduce the activation energy so the reaction can proceed faster. The reaction will proceed forward forming.

F. The following will favour CH4 at equilibrium

i. Catalyst to the reaction mixture,

ii. Both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture

iii. Adding more C to the reaction mixture.

4 0

3 years ago

I really need this fast

Sladkaya [172]

Answer:

drinka jila jokl ruhi

Explanation:

fihk gg secna isla

8 0

4 years ago

Read 2 more answers

Read the given chemical reaction.

k0ka [10]

It is d
D is for you

6 0

3 years ago

Read 2 more answers