A layer of sandstone is found on top of a layer of shale

Calculate the percentage by mass of chlorine in Cobalt (II) chloride (cocl2)

Likurg_2 [28]

Explanation:

Divide the mass of chlorine by the molar mass of cobalt chloride, then multiply by 100.

Molar Mass of Cobalt Chloride.

Mass of Chlorine in Cobalt Chloride.

Percent Composition of Chlorine.

8 0

3 years ago

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An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

Answer: The unknown salt is NaF

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

Initial: 0.1

At eqllm: 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

2 years ago

When the forces acting on the object are balanced the object will?

Volgvan

Answer:

When two forces acting on an object are of similar size but acting in opposite directions, we say they are forces of balance. If the forces on an object are balanced (or if there are no forces acting on it), this is what happens: the object stays stationary

8 0

2 years ago

Which formula shows a substance that is not molecular?

Scrat [10]

H ( hydrogen ) is the answer I believe.

8 0

3 years ago

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Consider the reaction of CaCN2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3 . How m

adelina 88 [10]

Answer:

56 g. Option 3.

Explanation:

The reaction is: CaCN₂ + 3H₂O → CaCO₃ + 2 NH₃

1 mol of calcium cianide reacts with 3 moles of water in order to produce 1 mol of calcium carbonate and 2 moles of ammonia

We have the mass of each reactant, so let's convert the mass to moles:

45 g. 1mol / 80.08 g = 0.562 moles of cianide

45 g. 1mol / 18 g = 2.5 moles of water

The cianide is the limiting reactant:

3 moles of water need 1 mol of cianide to react

Then, 2.5 moles of water will need (2.5 . 1)/ 3 = 0.833 moles

As we have 0.562 moles of CN⁻ we don't have enough

We can work now, on the reaction:

Ratio is 1:1. Therefore 0.562 moles of cianide will produce 0.562 moles of carbonate

Let's convert the mass to moles to find the answer:

0.562 mol . 100.08 g / 1 mol = 56.2 g

8 0

2 years ago