The ability to attract an electron for bonding is called:

What is the silver ion concentration in a solution prepared by mixing 369 mL 0.373 M silver nitrate with 411 mL 0.401 M sodium c

LuckyWell [14K]

Answer:

[A g + ] = 3.12 *10^-6 M

Explanation:

Step 1: Data given

Volume silver nitrate = 369 mL

Molarity silver nitrate = 0.373 M

Volume sodium chromate = 411 mL

Molarity sodium chromate = 0.401 M

The Ksp of silver chromate is 1.2 * 10^− 12

Step 2: The balanced equation

2 A g + ( a q ) + C rO4^2- ( a q ) →Ag2CrO4 ( s)

Step 3:

[Ag+]i = [AgNO3] * V1/(V1+V2) * 1 mol Ag+ / 1 mol AgNO3

[Ag+]i = 0.373 M * 0.369/ (0.369+0.411)

[Ag+]i = 0.176 M

[C rO4^2]i = [Ag2CrO4] * V2 /(V1+V2) * 1mol C rO4^2 / 1 mol Ag2CrO4

[C rO4^2i = 0.401 M * 0.411 / (0.369+0.411)

[C rO4^2]i = 0.211 M

Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^− 12

[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i

[CrO4^2-]f = 0.211 -0.088 = 0.123 M

1.2 * 10^− 12 = ( 2 x ) ²*( 0.123M + x )

[ C O 3^ −2] f >> x

1.2 * 10^− 12 = ( 2 x ) ²* 0.123M

x = 1.56 * 10^-6

[A g + ] f = 2x = 3.12 *10^-6 M

,

4 0

3 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

4 years ago

A chemist determines by measurements that 0.0200 moles of chlorine gas participate in a chemical reaction. Calculate the mass of

Schach [20]

Answer:

1.4200 g

Explanation:

The chlorine gas has a molecular formula Cl2. In table periodic, the molar mass of one atom of chlorine is 35.5 g/mol, so the molar mass of Cl2 is 71 g/mol.

Molar mass, mass and number of moles are related by the equation below:

$n = \frac{m}{MM}$

Where n is the number of moles, m is the mass, and MM is the molar mass of the compound.

So, for 0.0200 moles of Cl2 produced:

$0.0200 = \frac{m}{71}$

m = 71x0.0200

m = 1.4200

8 0

4 years ago

In the molecule Br2, how many bonds would be present between the two atoms of Br

Evgesh-ka [11]

Answer:

The correct answer is A) single

Explanation:

In the case of the Bromo atom, it requires 1 electron to complete its octet, therefore it shares 1 electron with the other Bromo atom.

8 0

4 years ago

A student is using a copper strip and a strong chlorine solution in her experiment. Which of these is a chemical property of one

Vsevolod [243]

A chemical property refers to that which cannot be physically determined. From the choices given above, the chemical property of one of the student's materials is the reactivity of the copper strip. Thus, the answer to this item is letter B.

6 0

3 years ago

Read 2 more answers