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Yakvenalex [24]
3 years ago
13

Consider the reaction when aqueous solutions of potassium hydroxide and ammonium nitrate are combined. The net ionic equation fo

r this reaction is:
Chemistry
1 answer:
Reil [10]3 years ago
6 0

Answer: OH^-(aq)+NH_4^+(aq)\rightarrow NH_3(g)+H_2O(l)

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

KOH(aq)+NH_4NO_3(aq)\rightarrow KNO_3(aq)+NH_3(g)+H_2O(l)

The equation can be written in terms of ions as:

K^+(aq)+OH^-(aq)+NH_4^+(aq)+NO_3^-(aq)\rightarrow K^+(aq)+NO_3^-(aq)+NH_3(g)+H_2O(l)

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

The ions which are present on both the sides of the equation are potassium and nitrate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is :

OH^-(aq)+NH_4^+(aq)\rightarrow NH_3(g)+H_2O(l)

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Benzene is an unsaturated molecule, but cyclohexane is saturated.
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What does it mean when water is supercooled?
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Which change is most likely to occur when a molecule of H2 and a molecule of I2 collide with proper orientation and sufficient e
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Read 2 more answers
What volume (L) of 0.250 M HNO3 is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water ?
IRINA_888 [86]
<h3>Answer:</h3>

1.75 L HNO₃

<h3>Explanation:</h3>

We are given;

Molarity of HNO₃ as 0.250 M

Mass of NaOH as 17.5 g

Volume of NaOH = 350 mL

We are required to calculate the volume of 0.250 M

We are going to first write the balanced reaction:

NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

Then, we calculate the number of moles of NaOH

Moles = Mass ÷ Molar mass

Molar mass of NaOH = 39.997 g/mol

          = 17.5 g ÷ 39.997 g/mol

          = 0.4375 moles

We can now calculate the number of moles of HNO₃ using the mole ratio from the equation;

Mole ratio of NaOH to HNO₃ is 1 : 1

Therefore, if moles of NaOH are 0.4375 moles then;

Moles of HNO₃ will also be 0.4375 moles

We can now calculate the volume of HNO₃

Morality = Number of moles ÷ Volume

Thus;

Volume = Number of moles ÷ Molarity

             = 0.4375 moles ÷ 0.250 M

             = 1.75 L

Therefore, the volume of HNO₃ is 1.75 L

5 0
3 years ago
How many moles of FeCI3 are present in 345.0 g FeCI3
Sphinxa [80]

Answer:

\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}

Explanation:

To convert from moles to grams, the molar mass must be used.

1. Find Molar Mass

The compound is iron (III) chloride: FeCl₃

First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).

  • Fe:  55.84 g/mol
  • Cl:  35.45 g/mol

There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.

  • FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol

This number tells us the grams of FeCl₃ in 1 mole.

2. Calculate Moles

Use the number as a ratio.

\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Multiply by the given number of grams, 345.0.

345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}

Flip the fraction so the grams of FeCl₃ will cancel.

345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}

345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }

\frac{345.0 \ mol \ FeCl_3}{162.19 }

Divide.

2.12713484 \ mol \ FeCl_3

3. Round

The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.

For the answer we calculated, that is the thousandth place.

The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

\approx 2.127 \ mol \ FeCl_3

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.

6 0
3 years ago
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