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Helen [10]
3 years ago
9

Consider an element Z that has two naturally occuring isotopes with the following percent abundances: the isotope with a mass nu

mber 22.0 is 65.0% abundant; the isotope with a mass number 24.0 is 35.0% abundant. What is the average atomic mass for element Z? Round your answer to the hundredth.
Chemistry
1 answer:
rodikova [14]3 years ago
3 0

Answer:

Z=22.70

Explanation:

It is given that,

An element Z that has two naturally occurring isotopes with the following percent abundances as follows :

The isotope with a mass number 22 is 65.0% abundant; the isotope with a mass number 24 is 35.0% abundant.

The average atomic mass for element Z is given by :

Z=\dfrac{22\times 65+24\times 35}{100}\\\\Z=22.7

So, the average atomic mass for element Z is 22.70.

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Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati
worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4  +  Cl-

Firstly, we calculate the number of moles of  the ammonia  as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

                 NH4+      H2O     ⇄  NH3        H3O+

I                0.242                           0             0

C                 -X                              +x              +X

E             0.242-X                          X              X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b    = K w

 

, where  K w- the self-ionization constant of water, equal to  

10 ^-14  at room temperature

This means that you have

K a = K w.K b   = 10 ^− 14 /1.8 * 10^-5 =  5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈  0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0
3 years ago
What is the freezing point of an aqueous 1.58 molal nacl solution?
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[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>an</em><em> </em><em>ester</em><em>,</em><em> </em><em>diethylether</em><em> </em><em>is</em><em> </em><em>formed</em><em>]</em>

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