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3 years ago

What is the overall equation for FeCl3(aq) + NH3(aq) + H2O(l)? the overall ionic equation? then the net ionic equation?

2 answers:

4 0

So the question ask to calculate the over all equation for a certain element and the following are
- over all equation - FE(OH)3+3NH4CI
- over all ionic equation - e(OH)3 (s) & 3 NH4+ (aq) & 3Cl- (aq) - net ionic equation - Fe+3+3OH-+3NH$++3CI-

Amanda [17]3 years ago

4 0

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of iron(III) chloride, ammonia and water is given as:

$FeCl_3(aq.)+3NH_3(g)+3H_2O(l)\rightarrow Fe(OH)_3(s)+3NH_4Cl(aq.)$

Ionic form of the above equation follows:

$Fe^{3+}(aq.)+3Cl-(aq.)+3NH_3(g)+3H_2O(l)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq.)+3Cl^-(aq.)$

As, chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and is the spectator ion.

The net ionic equation for the above reaction follows:

$Fe^{3+}(aq.)+3NH_3(g)+3H_2O(l)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq.)$

Hence, the net ionic equation is written above

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Partial charges will exist on each atom when a bond exists between two atoms with different electronegativities. Since the atom

lana66690 [7]

Answer:

negative, positive, increase

Explanation:

From the given question,

During the formation of bond, between two atoms with difference between their electronegativity-

The more electronegative atom, will pull the electrons towards itself , and hence acquires a partial negative charge,

And,

The less electronegative atom, will acquire a partial positive charge.

The more the difference between the electronegativity of the atoms, the more would be the magnitude of partial charge.

And, the less would be the difference between the electronegativity of the atoms, the lesser would be the magnitude of partial charge.

7 0

3 years ago

The equation to calculate mean rate of reaction g/s

tatyana61 [14]

amount of product formed or amount of reactants used / time

8 0

3 years ago

Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample

k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

Step 1: Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

Step 2: The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

Step 3: Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

Step 4: Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

Step 5: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

6 0

3 years ago

Chemicals are classified as health hazards when they pose which of the following hazardous effects?

AlexFokin [52]

Answer:

poisoning, breathing problems, skin rashes, allergic reactions, allergic sensitisation, cancer, and other health problems from exposure.

Explanation:

many hazardous chemicals are also classified as dangerous goods.

4 0

2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago