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Blababa [14]

Products are copper+ aluminium chloride
reactants are aluminium+copper chloride

5 0

3 years ago

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

2 years ago

What is the freezing point of a solution of 0.5 mol of LiBr in 500 mL of water? (Kf = 1.86°C/m) –1.86°C –7.44°C –5.58°C –3.72°C

tresset_1 [31]

Answer:

Last choice: - 3.72°C

Explanation:

The freezing point depression in a solvent is a colligative property: it depends on the number of solute particles.

The equation to predict the freezing point depression in a solvent is:

ΔTf = Kf × m × i

Where,

ΔTf is the freezing point depression of the solvent,

m is the molality,

Kf is the cryoscopic molal constant of the solvent, and i is the Van'f Hoff factor, which is the number of ions produced by each unit formula of the ionic compound.

The calcualtions are in the attached pdf file. Please, open it by clicking on the image of the file.

Download pdf

6 0

2 years ago

Please help this test ends in 30 min

RideAnS [48]

Answer:

Option C

CH₃CH₂CH₂COOH

Explanation:

Carbonxylic acids are compounds which has the general formula

R–COOH where R is an alkyl group.

Considering the options given in the question above,

For A:

CH₃CH₂OCH₂CH₃ is an ether compound with general formula ROR' where R and R' are both alkyl group.

For B:

CH₃CH₂CH₂CH₂OH is an alcohol with general formula ROH where R is an alkyl group.

For C:

CH₃CH₂CH₂COOH is a carbonxylic acid with general formula R–COOH where R is an alkyl group.

For D:

CH₃CH₂C=OCH₂CH₃ is a ketone compound with general formula RC=OR' where R and R' are both alkyl group

For E:

ClCH₂CH₂CH₂CH₂CH₂CH₂Br is simply an Alkyl halide with general formula XRX where X is an halogen (i.e F, Cl, Br or I) and R is an alkyl group.

From the above illustration, only option C contains a Carbonxylic compound.

8 0

2 years ago

A solution was prepared by mixing 50.0 g

frez [133]

Answer:

4.78 %.

Explanation:

mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.

mass % = (mass of solute/mass of solution) x 100.

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.

∴ mass % = (mass of solute/mass of solution) x 100 = (50.0 g/1047.0 g) x 100 = 4.776 % ≅ 4.78 %.

4 0

3 years ago