Why do elements in a vertical column of the periodic table have similar chemical properties?

Tentukan pH Mg(OH)2 jika ke dalam larutan MgCl2 0,2M ditambahkan NH4OH sehingga

Vadim26 [7]

Answer:

pH 8.89

Explanation:

English Translation

If the MgCl₂ solution of 0.2 M has its pH raised by adding NH₄OH, the precipitate will begin to form at a pH of approximately.

Given the solubility product (Ksp) of Mg(OH)₂ = 1.2 x 10⁻¹¹

Assuming all of the salts involved all ionize completely

MgCl₂ ionizes to give Mg²⁺ and Cl⁻

MgCl₂ ⇌ Mg²⁺ + 2Cl⁻

1 mole of MgCl₂ gives 1 moles of Mg²⁺

Since the concentration of Mg²⁺ is the same as that of MgCl₂ = 0.2 M

Mg(OH)₂ is formed from 1 stoichiometric mole of Mg²⁺ and 2 stoichiometric moles of OH⁻

Ksp Mg(OH)₂ = [Mg²⁺][OH⁻]²

(1.2 x 10⁻¹¹) = 0.2 × [OH⁻]²

[OH⁻]² = (6×10⁻¹¹)

[OH⁻] = √(6×10⁻¹¹)

[OH⁻] = 0.000007746 M

p(OH) = - log [OH⁻] = - log (0.000007746)

pOH = 5.11

pH + pOH = 14

pH = 14 - pOH = 14 - 5.11 = 8.89

Hope this Helps!!!

3 0

3 years ago

Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>

4 0

2 years ago

Compared to our solar system, the universe is

GuDViN [60]

Answer: Massive

Explanation: There are many galaxies out in the universe and it's possible they go on indefinitely. Out of all of these, our solar system is very very tiny. As an analogy, our universe would be like an atom which are the the smallest units of matter. There are many other galaxies that we just haven't been able to discover but they are there.

3 0

2 years ago

A device that does work in a circuit is called what ?

zavuch27 [327]

These are called resistors.

7 0

2 years ago

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Ammonia, NH 3 , may react with oxygen to form nitrogen gas and water. 4 NH 3 ( aq ) + 3 O 2 ( g ) ⟶ 2 N 2 ( g ) + 6 H 2 O ( l )

Alex Ar [27]

Answer:

The limiting reactant is NH₃

0.0186moles of N₂ are the one produced by the limiting reactant

0.020 moles of N₂ are the one produced by the reactant in excess

Explanation:

This is the reaction

4NH₃ + 3O₂ → 2N₂ + 6H₂O

We should calculate the moles of each reactant

Mass / Molar mass = Moles

3.55 g / 17g/m = 0.208 moles NH₃

5.33 g / 32g/m = 0.166 moles O₂

4 moles of ammonia react with 3 moles of oxygen

0.208 moles of ammonia react with (0.208 .3)/4 = 0.156 moles O₂

We have 0.166 moles of O₂ and we need 0.156 moles, so O₂ is the reactant in excess.

3 moles of O₂ react with 4 moles of NH₃

0.166 moles of O₂ react with (0.166 . 4)/ 3 = 0.221 moles

We have 0.208 moles NH₃ and we need 0.221, so NH₃ is the limiting reactant.

To know the moles of N₂, let's apply the Ideal Gas Law

P.V =n.R.T

1atm . 0.450L = n . 0.082 . 295K

0.450 / (0.082 .295) = 0.0186 moles

If we have 100 % yield reaction:

4 moles NH₃ make 2 moles N₂

0.208 moles NH₃ make (0.208 .2)/4 = 0.104 moles

So the % yield reaction is.

0.104 moles ___ 100%

0.0186 moles ___ 17.9%

0.0186moles of N₂ are the one produced by the limiting reactant.

3 moles of O₂ produce 2 moles N₂

0.166 moles O₂ produce (0.166 .2)/3 = 0.111 moles

Now, we apply the yield.

100% ____ 0.111 moles

17.9% = 0.020 moles

8 0

2 years ago

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