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Neko [114]
4 years ago
8

7. Phosphorus 32 ( 32), a radioactive isotope of phosphorus, has a half-life of 14.2 days. a. If the amount of radioactivity of

phosphorus 32 decays continuously at a constant rate, what is the decay constant? Round your answer to 4 decimal places. (10 marks) If 100 grams of this substance are present initially, determine the exponential decay function by using the decay constant found in part (a). (4 marks) c. What amount of radioactivity will be left after 6 days? Round your answer to 2 decimal places. (4 marks)
Chemistry
1 answer:
Wittaler [7]4 years ago
4 0

Explanation:

(a) Given that:

Half life = 14.2 days

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{14.2}\ days^{-1}

<u>The rate constant, k = 0.0488 days⁻¹ </u>

(b)Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration  = 100 g

So,  

[A_t]=100\times e^{-0.0488\times t}\ g

(c) Given t = 6 days

So,

[A_t]=100\times e^{-0.0488\times 6}\ g

<u>Amount left = 74.6171 g</u>

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inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
What is the formula or equation for<br><br>motion,speed and energy?
DedPeter [7]
Energy is equal to motion over speed
7 0
3 years ago
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Epsom salt is commonly purchased in the pharmacy for a variety of uses, anti-inflammatory, laxative, and cosmetic. epsom salt is
Dominik [7]
Epsom salt is MgSO4.
We assume x water of hydration in the crystalline form.

Molecular mass of MgSO4 = 24+32+4*16=120
Molecular mass of MgSO4.xH2O = 120+18x

By proportion, 
2.000/0.977 = (120+18x)/120
Cross multiply
0.977(120+18x) = 120*2.000
from which we solve for x
17.586x+117.24 = 240
x=122.78/17.586
=6.980

Answer: there are 7 water of hydration in Epsom salt, according to the experiment.

Note: more accurate (proper) results may be obtained by using exact values (3-4 significant figures) in the molecular masses.  However, since water of hydration is the nearest integer, using approximate values (to at least two significant figures) suffice.
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3 years ago
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How many kilograms of dry air are in a room that measures 15.0 ft by 18.0 ft by 8.00 ft? Use the density of air of 1.168 g/L. Th
Nady [450]

Answer:

0.0481 kg.

Explanation:

First, we need to find the volume of the room.

15 x 18 x 8 = 1350

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Now, we convert feet to centimeters

1350 x 30.48 = 41,148

The volume is 41,148 cm³

1 cm³ = 1 mL

So we can say that the volume of the room is also 41,148 mL

Then, we convert milliliters to liters

41,148 ÷ 1,000 = 41.148

There are 41.148 liters of air in the room

We multiply the air's density with volume in liters

41.148 x 1.168 = 48.060864

There are 48.060864 grams of air in the room

Finally, convert grams to kilograms

48.060864 ÷ 1,000 = 0.048060864

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Where can you find most of earth's freshwater?
rjkz [21]
The answer is A, underground
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