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Yanka [14]
3 years ago
8

Calculate the molar mass of the following:

Chemistry
1 answer:
Westkost [7]3 years ago
7 0

Answer:

a)C2HBrClF3 = 197.35 g/mol

b)C12H14N2CL2 = 229.06g/mol

c)C8H10N4O2 = 194.22g/mol

d) CO(NH2)2=60.07 g/mol

e)C17H35CO2Na = 306.52 g/mol

Explanation:

Molar mass of a compound is equal to the sum of the atomic masses of the constituent elements.

a) C2HBrClF3

Molar\ mass = 2(At. mass C)+1(at.mass H) +1(At. mass Br) + 1(At.mass Cl) + 3(At.mass F)\\=2(12.01 g/mol) + 1(1.01g/mol)+1(79.90 g/mol) +1(35.45g/mol)+3(18.99g/mol)=197.35g/mol

b) C12H14N2CL2

Molar\  mass = 12(C) + 14(H) + 2(N) + 2(Cl)\\\\=12(12.01) + 14(1.01) + 2(14.01) + 2(35.45) = 229.06g/mol

c) C8H10N4O2

Molar\  mass = 8(C) + 10(H) + 4(N) + 2(O)\\\\=8(12.01) + 10(1.01) + 4(14.01) + 2(16.00) =194.22g/mol

d) CO(NH2)2

Molar\ mass = 1(C) + 1(O) + 2(N) + 4(H)\\\\=1(12.01) + 1(16.00) + 2(14.01)+4(1.01) =60.07 g/mol

e) C17H35CO2Na

Molar Mass = 18(C) + H(35) +2(O) + 1(Na)\\\\=18(12.01) + 35(1.01) + 2(16.00) + 1(22.99) =306.52 g/mol

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Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.
Firlakuza [10]

Answer:

  • [Ag⁺] = 1.3 × 10⁻⁵M
  • s = 3.3 × 10⁻¹⁰ M
  • Because the common ion effect.

Explanation:

<u></u>

<u>1. Value of [Ag⁺]  in a saturated solution of AgCl in distilled water.</u>

The value of [Ag⁺]  in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                    0

C          -s                      +s                  +s

E         X - s                   s                     s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × s
  • Ksp = s²

By substitution:

  • 1.8 × 10⁻¹⁰ = s²
  • s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

  • [Ag⁺] = 1.3 × 10⁻⁵M ← answer

<u></u>

<u>2. Molar solubility of AgCl(s) in seawater</u>

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

  • AgCl(s)    ⇄    Ag⁺ (aq)    +    Cl⁻ (aq)

I            X                      0                  0.54

C          -s                      +s                  +s

E         X - s                   s                     s + 0.54

The new equation for the Ksp equation will be:

  • Ksp = [Ag⁺] [Cl⁻]
  • Ksp = s × ( s + 0.54)
  • Ksp = s² + 0.54s

By substitution:

  • 1.8 × 10⁻¹⁰ = s² + 0.54s
  • s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

     

     s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

<u>3. Why is AgCl(s) less soluble in seawater than in distilled water.</u>

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

3 0
3 years ago
A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq )
Alex73 [517]

Answer:

0.00369 moles of HCl react with carbonate.

Explanation:

Number of moles of HCl present initially = \frac{0.1200}{1000}\times 50.0 moles = 0.00600 moles

Neutralization reaction (back titration): NaOH+HCl\rightarrow NaCl+H_{2}O

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration = \frac{0.0980}{1000}\times 23.60 moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.

3 0
3 years ago
If the density of an object is 5.2g/cm3, and it’s volume is 3.7 cm3, what is it’s mass?
lord [1]
Here's the equation you use: Density = mass/volume 

1) 5.2g/cm^3 = m/3.7cm^3 

2) m = 5.2g/cm^3 x 3.7cm^3 

3) m = 19.24g 

You can check the answer by plugging it in 

19.24g/3.7cm^3 
= 5.2g/cm^3
5 0
3 years ago
PO3−4 what's the formula name?
masha68 [24]

Answer:

Phosphate ion

Explanation:

4 0
3 years ago
How many moles of nitrogen gas are there in 16.8 L of this gas at STP?
sweet-ann [11.9K]
According to Avogadro's Law, same volume of any gas at standard temperature and pressure will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).

Data Given:

                  n = moles = ?
                  V = Volume = 16.8 L

Solution:
               
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....

                           = ( 1 mole × 16.8 L) ÷ 22.4 L
                    
                           = 0.75 moles

Result:
           
16.8 L of Nitrogen gas will contain 0.75 moles at standard temperature and pressure.
4 0
3 years ago
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