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sukhopar [10]
3 years ago
10

Analysis of an athletes urine found the presence of a compound with a molar mass of 312 g/mol. How many moles of this compound a

re contained in 30.0 mg? How many molecules of the compound is this?
Chemistry
1 answer:
rewona [7]3 years ago
5 0
<h3>Answer:</h3>

= 5.79 × 10^19 molecules

<h3>Explanation:</h3>

The molar mass of the compound is 312 g/mol

Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)

We are required to calculate the number of molecules present

We will use the following steps;

<h3>Step 1: Calculate the number of moles of the compound </h3>

Moles=\frac{mass}{molar mass}

Therefore;

Moles of the compound will be;

=\frac{0.030}{312g/mol}

      = 9.615 × 10⁻5 mole

<h3>Step 2: Calculate the number of molecules present </h3>

Using the Avogadro's constant, 6.022 × 10^23

1 mole of a compound contains 6.022 × 10^23  molecules

Therefore;

9.615 × 10⁻5 moles of the compound will have ;

= 9.615 × 10⁻5 moles × 6.022 × 10^23  molecules

= 5.79 × 10^19 molecules

Therefore the compound contains 5.79 × 10^19 molecules

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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

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