Is ammonia a compound or an element?

What is the freezing point depression of a solution that contains 0.705?

Sveta_85 [38]

Colligative properties calculations are used for this type of problem. Calculations are as follows:

ΔT(freezing point) = (Kf)m
ΔT(freezing point) = 1.86 °C kg / mol (0.705)
ΔT(freezing point) = 1.3113 °C

Hope this answers the question. Have a nice day.

6 0

2 years ago

Which of the following is true?

Tomtit [17]

Answer:

a. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

Explanation:

The change in free energy (ΔG) that is, the energy available to do work, of a system for a constant-temperature process is:

$ΔG = ΔH - TΔS$

When ΔG < 0 the reaction is spontaneous in the forward direction.

When ΔG > 0 the reaction is nonspontaneous. The reaction is

spontaneous in the opposite direction.

When ΔG = 0 the system is at equilibrium.

If both ΔH and ΔS are positive, then ΔG will be negative only when the TΔS term is greater in magnitude than ΔH. This condition is met when T is large.

3 0

2 years ago

Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 162. kJ during a certain chemical reaction, w

puteri [66]

Answer:

= -356KJ

therefore, the reaction where heat is released is exothermic reaction since theΔH is negative

Explanation:

given that enthalpy of gaseous reactants decreases by 162KJ and workdone is -194KJ

then,

change in enthalpy (ΔH) = -162( released energy)

work(w) = -194KJ

change in enthalpy is said to be negative if the heat is evolved during the reaction while heat change(ΔH) is said to be positive if the heat required for the reaction occurs.

At constant pressure the change in enthalpy is given as

ΔH = ΔU + PΔV

ΔU = change in energy

ΔV = change in volume

P = pressure

w = -pΔV

therefore,

ΔH = ΔU -W

to evaluate energy change we have,

ΔU =ΔH + W

ΔU = -162+ (-194KJ)

= -356KJ

therefore, the reaction where heat is released is exothermic reaction since theΔH is negative

6 0

2 years ago

In the following reaction, how many grams of ammonia (NH3) will react with 27.8 grams of nitric oxide (NO)? 4NH3 + 6NO → 5N2 + 6

Rzqust [24]

4NH₃ + 6NO → 5N₂ + 6H₂O

mol of NO = $\frac{mass of NO}{Molar Mass of NO}$

= $\frac{27.8 g}{30.01 g / mol}$

 = 0.93 mol

Based on the balance equation mole ratio of NH₃ : NO is 4 : 6
= 2 : 3

If mol of NO = 0.93 mol

then mol of NH₃ = $\frac{0.93 mol * 2}{3}$

= 0.62 mol

Mass of ammonia = mol × molar mass

= 0.62 mol × 17.03 g/mol

= 10.54 g

Therefore B is the best answer

3 0

2 years ago

Read 2 more answers

5. Complete and balance an equation for each reaction:

Kipish [7]

CaI₂ + Hg(NO₃)₂ --------->HgI₂ + Ca(NO3)2

2Al + 3Cl₂ --------->2AlCl3

Ag + HCl ------->AgCl + H2

C2H2 + 5O2 --------> 4CO2 + 2H2O

MgCl₂ --------->Mg + Cl2

6 0

2 years ago