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3 years ago

Which colligative property is employed when salt is put on an icy sidewalk to melt ice?

2 answers:

6 0

<span>For this number, we are to determine the colligative property that is affected by the addition of salt in the sidewalk in order to melt the ice. The answer is that the colligative property being displayed is the freezing point depression. Colligative properties are innate properties of substances and are only dependent on the amount of substance being added. </span>

ankoles [38]3 years ago

4 0

<span>The correct answer is 'freezing point depression'. Colligative properties depend on the concentration of molecules of a solute. Examples of other colligative properties are boiling point elevation or vapour pressure lowering. The salt causes ice on the side walk to melt because it lowers the freezing point. </span>

You might be interested in

CO2(g) + H2O(I) - -> C6H12O6(s) + O2(g)

Lana71 [14]

Answer:

For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but PhC2H5 + O2 = PhOH + CO2 + H2O will; Compound states [like (s) (aq) or (g)] are not required. If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.

Explanation:

4 0

2 years ago

You make a solution by putting 45.6g of iron lll carbonate into 167ml of water. What is it's molarity?

ludmilkaskok [199]

1. The molar mass of Fe2(CO3)3 is 291.72 g/mol. This means that 45.6 g is equivalent to 0.156 mol. Dividing by the 0.167 L of water gives a solution of 0.936 M.
2. Multiplying (0.672 M)(0.025 L) = 0.0168 mol. The molar mass of Ni(OH)2 is 92.71 g/mol, so multiplying by 0.0168 mol = 1.56 grams. Therefore you would need to dissolved 1.56 g of Ni(OH)2 into 25 mL of water.
3. Fe2(CO3)3 + Ni(OH)2 --> Fe(OH)3 + NiCO3Balancing: Fe2(CO3)3 + 3Ni(OH)2 --> 2Fe(OH)3 + 3NiCO3The reaction quotient is:[Fe(OH)3]^2 * [NiCO3]^3 / [Fe2(CO3)3][Ni(OH)2]^3= (0.05)^2 * (1.45)^3 / (0.936)(0.672)^3= 0.0268Since this is < 1, it implies that the reactants are favored at equilibrium.

4 0

3 years ago

If 5.0 g of copper metal reacts with a solution of silver nitrate, how many grams of silver metal are recovered?

sukhopar [10]

Answer:

16.9g

Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2

Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.

Then to calculate the theoretical yield you need to compare moles of the reactants:

m(Cu)=5g

M(Cu)=63.55

n(Cu)=5/63.55=0.0787

By comparing coefficients you require twice as much silver: 0.157mol

n(Ag)=0.157

M(Ag)=107.86

m(Ag)=0.157x107.86=16.9g

Hence, the theoretical yield of this reaction would be 16.9g

3 0

2 years ago

ammonia (NH3(g) Hf=-45.9 kJ/mol) reacts. with oxygen to produce nitrogen and water (H2O(g) Hf = -241.8 kJ/mol according to the e

kherson [118]

Answer:

ΔH°_rxn = -195.9 kJ·mol⁻¹

Explanation:

4NH₃(g) + 3O₂(g) ⟶ 2N₂(g) +6H₂O(g)

ΔH°_f/(kJ·mol⁻¹): -45.9 0 0 -241.8

The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is

ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)

ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ

ΣΔH°_f(reactants) = -4(45.9) = -183.6 kJ

ΔH°_rxn = (-1450.8 + 183.6) kJ = -1267.2 kJ

6 0

2 years ago

Read 2 more answers

A solution of barium nitrate has 61.2g of barium nitrate in 1 liter of solution. How many mg of barium are there in 7.5 quarts

Nuetrik [128]

Answer:

220.44g Ba²⁺ ions in solution

Explanation:

Given parameters:

Mass of barium nitrate = 61.2g

Volume of solution = 1 liter

Unkown:

Mass of barium in 7.5quarts of solution?

Solution

We must first convert quarts to its liter equivalence:

1 quarts = 0.95 liter

7.5 quarts = 0.95 x 7.5; 7.125liter

Now, let us find the mass of barium nitrate in a solution of 7.125liter:

Given:

1 liter of solution contains 61.2g of barium nitrate:

7.125 liter will contain 7.125 x 61.2 = 436.05g of barium nitrate.

The formula of the compound is Ba(NO₃)₂:

In solution we have Ba²⁺ + NO₃⁻

Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻

Number of moles of Ba(NO₃)₂ = $\frac{mass of Ba(NO₃)₂}{Molar mass of Ba(NO₃)₂}$

Molar mass of Ba(NO₃)₂ = 137 + 2[14 + 3(16)] = 271g/mol

Number of moles of Ba(NO₃)₂ = $\frac{436.05}{271}$ = 1.609mole

1 mole of Ba(NO₃)₂ will produce 1 mole of Ba²⁺ ions in solution

therefore, 1.609mole of Ba(NO₃)₂ will also yield 1.609mole of Ba²⁺ ions in solution

Mass of Ba²⁺ ions = Number of moles of Ba²⁺ ions x molar mass of Ba²⁺ ions

Mass of Ba²⁺ ions = 1.609 x 137 = 220.44g

3 0

3 years ago