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Iteru [2.4K]
3 years ago
6

Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?

Chemistry
1 answer:
Rufina [12.5K]3 years ago
6 0
Radioactive decay => C = Co { e ^ (- kt) |

Data:

Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min

Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

=> k = ln {Co / C} / t =  ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719

2) Use C / Co  = 1/2 to find the hallf-life

C / Co = e ^ (-kt) => -kt = ln (C / Co)

=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k

t = ln(2) / 0.44719 = 1.55 hr.

Answer: 1.55 hr
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What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. m
r-ruslan [8.4K]

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2


Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol                               2 mol
0.100 mol                           0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.



4 0
3 years ago
43 mg = [?]g <br>A. 0.043 g <br>B. 4.3 g <br>C. 4300 g <br>D. 43,000 g​
Ymorist [56]

Answer:

Option A (0.043 g) is the correct answer.

Explanation:

Given:

= 43 mg

As we know,

1 \ mg = \frac{1}{1000} \ g

then,

⇒ 43 \ mg = \frac{43}{1000}  \ g

              = 0.043 \ g

Thus, the above is the correct alternative.

4 0
3 years ago
What is the relationship between radius and diameter
Mrrafil [7]

Answer:

the radius is half the diameter

6 0
2 years ago
Read 2 more answers
The air in a 2 L balloon at 0.998 atm and 34.0 °C. What will be its pressure if it is brought to a higher altitude where it now
Phantasy [73]

Answer: 0.529 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.998 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 2 L

V_2 = final volume of gas = 3.5 L

T_1 = initial temperature of gas = 34.0^oC=273+34.0=307.0K

T_2 = final temperature of gas = 12.0^oC=273+12.0=285.0K

Now put all the given values in the above equation, we get:

\frac{0.998\times 2}{307.0K}=\frac{P_2\times 3.5}{285.0K}

P_2=0.529atm

Thus the pressure if it is brought to a higher altitude where it now occupies 3.5 L and is at 12.0 °C is 0.529 atm

4 0
3 years ago
Which is not an indicator for a chemical change?
Anastasy [175]

Answer:

A

Explanation:

Because it does not change entirely.

8 0
3 years ago
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