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garri49 [273]
3 years ago
9

After decaying for 48 hours, l/16 of the original mass of a radioisotope sample remains unchanged. What is the half-life of this

radioisotope?
Chemistry
1 answer:
ivolga24 [154]3 years ago
3 0

After the first half-life, 1/2 is unchanged. After the 2nd one, 1/4 remains. After the 4th one, 1/16 remains. So you have 4 half-lives in 48 hours, the half-life is thus 12 hours.
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What is the difference between an atom and an element?
Travka [436]

Answer:

A particular atom will have the same number of protons and electrons and most atoms have at least as many neutrons as protons. An element is a substance that is made entirely from one type of atom

Explanation:

so In simple terms an atom contains Protons and Electrons and can be found in an Element, but an element comprises of atoms which come together to form up that particular Element

4 0
3 years ago
How much heat is required to increase the temperature of 10.0 grams of water 6.0oC? (The specific heat of water is 4.18 J/g x oC
scoray [572]

Heat required in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC (T2-T1)
Heat = 10.0 g (4.18 J/g-C ) ( 6.0 C )
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8 0
3 years ago
Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constan
Valentin [98]

Answer : The equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

Explanation :

The given main chemical reaction is:

N_2(g)+2O_2(g)\rightarrow N_2O_4(g);  K

The intermediate reactions are:

(1) N_2O_4(g)\rightarrow 2NO_2(g);  K_a

(2) \frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g);  K_b

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) 2NO_2(g)\rightarrow N_2O_4(g);  \frac{1}{K_a}

(2) N_2(g)+2O_2(g)\rightarrow 2NO_2(g);  (K_b)^2

Thus, the equilibrium constant for this reaction will be:

K=\frac{1}{K_a}\times (K_b)^2

K=\frac{(K_b)^2}{K_a}

Thus, the equilibrium constant for this reaction is, K=\frac{(K_b)^2}{K_a}

5 0
3 years ago
What is the density of 1.00 mol of carbon monoxide gas at STP?
Ksju [112]

Answer:

Density = mass / volume.

If this is an ideal gas then 1mol will take up 22.4L of volume (fact about ideal gases you should remember)

Since you have 1mol then you know the volume of the gas that you have (22.4L)

Now, just convert 1mol of CO to grams.The ptable tells you that the mass of 1 mol of C is 12g and the mass of 1 mol of O is 16g. So the mass of 1 mol of CO is......... :):)

Now you have the mass and the volume, so just divide :) enjoy

6 0
2 years ago
Consider the rate constants below. which would correspond to the highest order reaction? k = 8.79 x 102 ms–1 k = 4.46 x 10–1 m–1
Kipish [7]
When the order n of the reaction rate is determined by this formula:

rate = K * [A]^n

when we have the unit of rate M S^-1 

and the unit of [A] M

So:

1- for k = 8.79 x 10^2 M S^-1 :

when the unit of K here is M S^-1

by substitution in rate formula:

M S^-1 = M S^-1 * M ^n

∴ M^n = 1

∴M^0 = 1 

∴ this reaction rate is zero order reaction


2- when k = 4.46 x 10^-1  and has unit M-1 S-1:

by substitution in rate formula:

M S^-1 = M-1 S-1  * M^n

∴M^2 = M^n

∴n = 2 so this is the second order reaction

3- when K 2.35 x 10^6 and it's unit is S-1 :

So by substitution in rate formula:

M S^-1 = S^-1 * M^n

∴ M^n = M

∴ n = 1 

So this is the first order reaction

4- when k = 1.88 x 10^-3 and its unit is M-2 S-1 

by substitution in rate formula:

M S^-1 = M^-2 S^-1 * M^n

∴ M^n = M^3

∴ n = 3 

∴ this is a third order reaction 

from 1 & 2 & 3 & 4 

so we can see that the correct answer is (4), K= 1.88 x 10^-3 M-2 S-1 is the highest order reaction.


3 0
3 years ago
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