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Shtirlitz [24]
3 years ago
13

Which branch of chemistry studies the composition of substances

Chemistry
1 answer:
emmainna [20.7K]3 years ago
8 0

Analytical

Analytical chemistry seeks to determine the composition of substances.

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II. Ionic Equations
mario62 [17]

Answer:

Complete ionic: \begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Net ionic: \begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

Explanation:

Start by identifying species that exist as ions. In general, such species include:

  • Soluble salts.
  • Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2. These three salts will exist as ions:

  • Each \rm AgNO_3\, (aq) formula unit will exist as one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion.
  • Each \rm CaCl_2 formula unit will exist as one \rm Ca^{2+} ion and two \rm Cl^{-} ions (note the subscript in the formula \rm CaCl_2\!.)
  • Each \rm Ca(NO_3)_2 formula unit will exist as one \rm Ca^{2+} and two \rm {NO_3}^{-} ions.

On the other hand, \rm AgCl is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite \rm AgNO_3, \rm CaCl_2, and \rm Ca(NO_3)_2 (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each \rm AgNO_3\, (aq) formula unit will exist as only one \rm Ag^{+} ion and one \rm {NO_3}^{-} ion. However, because the coefficient of \rm AgNO_3\, (aq)\! in the original equation is two, \!\rm AgNO_3\, (aq) alone should correspond to two \rm Ag^{+}\! ions and two \rm {NO_3}^{-}\! ions.

Do not rewrite the salt \rm AgCl because it is insoluble.

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}.

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of \rm Ca^{2+} and two units of \rm {NO_3}^{-}. Doing so will give:

\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}.

Simplify the coefficients:

\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}.

7 0
3 years ago
Velocity is a vector quantity.Why?​
Alika [10]

Answer:

Has size and magnitude whereas a scalar quantity has only size

3 0
3 years ago
How many valence electrons does Nitrogen have?​
irinina [24]

Answer:

5 valence electrons

Nitrogen has a total of 5 valence electrons

Explanation:

6 0
3 years ago
What is FeBr3 compound name?
inessss [21]
Iron bromide. Iron bromide (FeBr3)
3 0
3 years ago
Read 2 more answers
Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
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