1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nina [5.8K]
3 years ago
7

A 40.0 milligram sample of 33p decays to 10.0 milligrams in 50.0 days. What is the half life of 33p?

Chemistry
1 answer:
il63 [147K]3 years ago
8 0

Answer:

half life = 25 days

Explanation:

Given data:

Total amount of sample = 40 mg

Sample left after 50 days = 10 mg

Half life of P³³ = ?

Solution:

<em>1st step:</em>

when time is zero = 40 mg

1 half life = 40/2 = 20 mg

2 half life = 20/ 2= 10 mg

<em>2nd step:</em>

<em>Elapsed time/ half life = HL</em>

half life = 50 days/ 2

half life = 25 days

You might be interested in
CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

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

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

7 0
3 years ago
Plants such as _______________ provided much nitrogen to the soil.
Allushta [10]

Answer:

Abiotic

Explanation:

Its because it includes water, sunlight, tempature and soil

6 0
2 years ago
(d)
Masteriza [31]

Explanation:

hope this helps you,

-s.

4 0
2 years ago
If a car travels 400 meters in 20 sec how fast is it going
pochemuha

Answer:

400 meters every 20 seconds

Explanation:

7 0
3 years ago
Read 2 more answers
How to compare and contrast a solution of hydrogen chloriide gas in water and toluene?
Airida [17]
<span>The solute is the substance that is being dissolved while the solvent is the base that the solute is bring dissolved in. For example, in salt water, salt would be the solute that dissolves into the water, and the water is the solvent that the salt is being dissolved in.</span>
4 0
3 years ago
Other questions:
  • Which is an element? A) alcohol B) sodium C) sugar D) water
    12·2 answers
  • A parchment fragment was discovered that had about 68% as much 14C radioactivity as does plant material on Earth today. Estimate
    6·1 answer
  • The inca "City of the sun"was ​
    6·1 answer
  • All of the elements below can exist as network solids EXCEPT 1. As 2. B 3. Si 4. O 5. C
    7·1 answer
  • If 75 kJ of heat is transferred to 1200 g liquid water at 36°C, what would the
    10·1 answer
  • Please help ASAP PLEASE
    9·1 answer
  • Balance the reaction and determine which of the following coefficients
    7·1 answer
  • Is CH4O2 polar or nonpolar?
    9·1 answer
  • A student has a balloon with a volume of 2.5 liters that contains 4.0 moles of air. The ballon has a small leak, allowing one mo
    9·1 answer
  • WILL MARK AS BRAINLIEST IF CORRECT
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!