What is the identify of substance X if 0.380 mol of X weighs 17.5 g? A) NO_2 B)

Question

What is the identify of substance X if 0.380 mol of X weighs 17.5 g? A) NO_2 B) NO_3 C) N_2O D) N_2O_4 How many grams of calcium chloride are needed to produce 10.0 g of potassium chloride? CaCl_2(aq) + K_2CO_3(aq) rightarrow 2 KCI(aq) + CaCO_3(aq) A) 0.134 g B) 7.44 g C) 14.9 g D) 29.8 g How many molecules of N_2 are in a 500.0 mL container at 780 mm Hg and 135 degree C? A) 8.76 times 10^21 B) 9.23 times 10^21 C) 2.65 times 10^22 D) 2.79 times 10^22 An "empty" aerosol can at 25 degree C still contains gas at 1.00 atmosphere pressure. If an empty can is thrown into a 525 degree C fire, what is the final pressure in the heated can? A) 2.68 atm B) 0.398 atm C) 2.51 atm D) 19.0 atm At STP how many liters of NH_3 can be produced from the reaction of 6.00 mol of N_2 with 6.00 mol of H_2? N_2(g) + 3H_2(g) rightarrow 2NH_3(g) A) 44.8 L B) 89.6 L C) 134 L D) 269 L Of the following, which element has the highest first ionization energy? A) beryllium B) boron C) hydrogen D) lithium

Explanation / Answer

(45)

Moles = mass / molar mass

molar mass = 17.5 / 0.380 = 46.0

(A) NO2 has molar mass of 46 g/mol

(46)

2 mol of KCl = 1 mol CaCl2

Then, 10.0 / 74.5 mol of KCl = 10.0 / (74.5 * 2) = 0.0671 mol of CaCl2

So, Mass of CaCl2 = 0.0671 * 111 = 7.44 g.

(B)

(47)

V = 500 mL = 0.500 L

P = 780 / 760 atm = 1.03 atm

T = 135 + 273.15 = 408.15 K

R = 0.0821 L.atm.K-1.mol-1

V = ?

Ideal gas equation,

P V = n R T

n = 1.03 * 0.500 / (0.0821 * 408.15) = 0.0154 mol

1 mol of any species = 6.022 * 1023 particles

then, 0.0154 mol = 0.0154 * 6.022 * 1023 = 9.26 * 1021

(B)

(48)

According to gye Lussac's law,

P1/T1 = P2/T2

1.00 / 298.15 = P2 / 798.15

P2 = 2.68 atm

(A)

(49)

N2 is limiting reagent.

1 mol N2 can form 2 mol NH3

then, 6 mol N2 can form 12 mol NH3

At STP 1 mol of any gas occupies 22.4 L

Then, 12 mol of NH3 gas occupies 12 * 22.4 = 268.8 L

(D)

(50)

(C) As it smaller in size.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.