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3 years ago

30 points if you can help me with this chemistry question

Chemistry

1 answer:

PIT_PIT [208]3 years ago

5 0

answer:

i am pretty sure you use cross charges for this.

so i think it is the third choice.

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Which of the following can be said in the crystallization of liquid salt. A.

forsale [732]

Answer:

Correct option is D)

Explanation:

Crystallization is the process that is used to separate a pure solid in the form of its crystals from a solution.

Copper sulphate, salt, and potash alum can be purified by this method. In this process, the solution of the substance is heated and then cooled to form crystals.

8 0

2 years ago

How many unpaired electrons are present in the ground state of an atom from each of the following groups?

Paha777 [63]

Explanation:

General electronic configuration for the given groups are as follows.

For 4A(14) : $ns^{2}np^{4}$

This means that there are 4 unpaired electrons in group 4A(14).

For 7A(17) : $ns^{2}np^{5}$

So, there are 7 unpaired electrons present in group 7A(17).

For 1A(1) : $ns^{1}$

Hence, there is only one unpaired electron present in group 1A(1).

For 6A(16) : $ns^{2}np^{4}$

So, there are 6 unpaired electrons present in group 6A(16).

3 0

4 years ago

Plz answer these questions :

SSSSS [86.1K]

Answer: 2) see below

3) 2AgCl → 2Ag + Cl₂

Explanation:

Step 1: Crushing & Grinding CuFeS₂(s)→ CuFeS₂(s)

Step 2: Froth Flotation CuFeS₂(s)----> CuFeS₂ (l)

Step 3: Roasting 2CuFeS₂(l) + 3O₂(g) → 2FeO(s) + 2CuS(s) + 2SO₂(g)

Step 4: Converting matte to blister Cu₂S(l) + O₂(g) → 2Cu(l) + SO₂(g)

Step 5: Anode Casting Cu(l) → Cu(s)

Step 6: Electrolytic Refining Cu(s) → Cu(s)

Anode: Cu(s) → Cu₂ + (aq) + 2e-

Cathode: Cu2 + (aq) + 2e- → Cu(s)

see diagram below for illustration of steps

3) When silver chloride (solid) is exposed to sunlight, it decomposes into silver (solid) and chlorine (gas).

The equation is written as: $2AgCl_{(s)}\xrightarrow{\text{sunlight}}}2Ag_{(s)}+Cl_{2(g)}$

3 0

4 years ago

What is work and why energy is useful for work??? (please guys give answer)

dexar [7]

Work is labor and energy is useful for it because you need energy to work unless you will be tired

3 0

3 years ago

If the [H+] in a solution is 1 × 10–1 mol/L, what is the [OH–]? Show your work.

AVprozaik [17]

Hello!

The basic equations to solve this is
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
------------------------------------------------------------------------------------------------------
Find pH

pH = -log(1 * 10^-1)
pH = 1
------------------------------------------------------------------------------------------------------
Find pOH
1 + pOH = 14

pOH = 13
------------------------------------------------------------------------------------------------------
Find OH-
[OH-] = 10^(-pOH)
[OH-] = 1 * 10^-13mo/L

The answer is $[OH-] = 1 * 10^{-13} mol/L$

Hope this helps!

3 0

4 years ago