Chemistry, 05.07.2019 10:30 princessa15266

Balance the following reaction. a coefficient of "1" is understood. choose option "blank" for the correct answer if the coefficient is "1." koh + cu(no3)2 kno3 + cu(oh)2 a) blank b) 2 c) 3 d) 4

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Answer from: bbykk6023

2KOH + [blank]Cu(NO₃)₂⟶ 2KNO₃ + [blank]Cu(OH)₂

Explanation:

Step 1. Start with the most complicated-looking formula [Cu(NO₃)₂].

Put a 1 in front of it and balance its atoms.

KOH + 1Cu(NO₃)₂⟶ KNO₃ + Cu(OH)₂

Step 2. Balance Cu.

We have fixed 1 Cu on the left. We need 1 Cu on the right. Put a 1 in front of Cu(OH)₂.

KOH + 1Cu(NO₃)₂⟶ KNO₃ + 1Cu(OH)₂

Step 3. Balance N.

We have fixed 2 N on the left-hand side. We need 2 N atoms on the right. Put a 2 in front of KNO₃.

KOH + 1Cu(NO₃)₂⟶ 2KNO₃ + 1Cu(OH)₂

Step 4: Balance O

We have fixed 8 O on the right and 6 O on the left. We need 2 more O atoms on the left.Put a 2 in front of KOH.

2KOH + 1Cu(NO₃)₂⟶ 2KNO₃ + 1Cu(OH)₂

Every formula now has a coefficient. The equation should be balanced. Let’s check.

Atom On the left On the right

K 2 2

N 2 2

Cu 1 1

O 8 8

All atoms are balanced.

The balanced equation is

2KOH + [blank]Cu(NO₃)₂⟶ 2KNO₃ + [blank]Cu(OH)₂

Answer from: Quest

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.

Answer from: Quest

the dissociation of nickel hydroxide is follows:

$ni(oh)_2 > ni^2^+ + 2 oh^-$

$ksp for no(oh)2 = 5.5 * 10^-^1^6$

$5.5 * 10^-^1^6 = s * (2 s)^2 = 4 s^3$

$s^3 = 137.5 * 10^-^1^8$

$s = 5.14 * 10^-^6 = [oh^-]$

the concentration of hydroxide ion is

$5.14 * 10^-^6 = [oh^-]$

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

$k = [ ni^2^+] * [oh^-]^2$

$6.1 * 10^-^1^6 = [ ni^2] * [5.14 * 10^-^6]^2$

$[ ni^2^+] = 0.23 * 10^-^4 mol /l$

thus, concentration of ni is 0.23 * 10^-^4 mol /l

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