Consider the reaction: 3 H2(g) + N2(g) ® 2 NH3(g). For a mixture of 2.00 mol N2

2. 1 mol C. 1mol CO. ×. 12.011g C. 1 mol C. = 0.6116 g C. One point is earned for

Regents Prep: Chemistry Multiple-Choice Questions .

A.0.011 mol L. -1 s. -1. B.0.019 mol L. -1 s. -1. C.0.044 mol L. -1 s. -1. D.0.049 mol

(a) 2N0(2)(g) vv N(2)O(4)(g). (b) NH(3)(g) + Hcl(g) vv NH(4)Cl(s). For both cases )

Suppose we place a mixture of 2.00 mol of H2, 1.00 mol of N2, and 2.00 mol of

158.53 g mol–1 SrCl. 2. = 9.16 g SrCl. 2. We would need an equal number of

Consider the following equilibrium: N2(g) + O2(g) ⇄ 2NO(g) Keq = 0.010. Initially,

1 mole H. 2. S = ? grams H. 2. S b. ? grams AgNO. 3. = 1 mole AgNO. 3.

(possible cathode reaction) 2 H2O (l) + 2 e- --> H2 (g) + 2OH-(aq) Eored = -0.83

MM of C4H10O2 = 4(12.0) + 10(1.0) + 2(16.0) = 90.0g/mol . . n = number of

b) M = (2 mol N) (14.01 g N/mol N) + (3 mol O) (16.00 g O/mol O) = 76.02 g/mol of

4.00 g Fe x = 7.16 x 10–2 mol Fe. Converting kilograms to moles: kg -> g ->

(2.) Water (3.00 mol) (3.) Ethanol (100 cm3) (4.) Nitrogen (25.0 litre). 6, The price

Hint: calculate total moles (n), subtract moles of H2 from n to get moles of He;

Ptotal = P1 + P2 + . . . Pn. Pt is the total pressure of a sample which contains a . .

2. 3 8 2. 1 mol N. 313 g N. 0.0313 g N. 0.002235 mol N. 14.0067 g N. 1 mol O g O

mol of N2O4. limiting mol N2. g N2. multiply by M. Sample Problem 3.10 . 3 mol

Ex. Molar mass of CaCl2; Avg. Atomic mass of Calcium = 40.08g; Avg. . 40.08 g/

6 Li (s) + 1 N2 (g) à 2 Li3N (s) (unbalanced) 0.305 mol N2. 4. An orange-brown

1mol of N=14gr 2mol =28gr 1mol of Cl=6 x 10^23 atom Cl 3.01 x 10^23 atoms Cl=.502mol 0f Cl 1mol Cl=35.45gr .502 mol Cl=17.80gr

n = 0.0783 mol O2 were produced n=0.157 mol . of moles/liter. If we know the

0.828 mol N/0.828 = 1.00 mol N. 4.13 mol O/0.828 = 4.99 mol O. Empirical

(4.0 M)(0.045 L) = 0.18 moles Na3PO4. 10. What is the volume of a balloon filled

yoctomolar, yM, 10−24 mol/dm3 . where NA is the Avogadro constant,

Nitrogen has 14.01 grams per mole. Therefore, 2 mole N has 2 * 14.01, or 28.02,

Calculate the maximum amount, in grams, of HgO formed. . Determine the mol

What is the total pressure exerted by a mixture of 2.00 g of H2 (MW = 2.016 g/mol

62.7 g H2SO4 Ũ [1 mol H2SO4/98.1 g] Ũ [2 mol Al/3 mol H2SO4] Ũ [26.98 g Al/1

28.0 g x. 1 mol. 2. = 1.72 g N b) How much heat is transferred when 25.0 g of

A mixture of 2 moles of O2 (g) and 1 mole of N2 (g) is maintained in a rigid

For example, if you are given 43.89% C, then use at least 12.01g/mol for the

N mole 1. Mg mole 3 g. 28.0134. N mole 1. N g 5.47. 2. 2. 2. = χ χ χ. 21. The

2. 132.03 g. 3. 1.99 mol. 4. 4240 g. 5. Calculate moles in 168.0 g of HgS. 6. . of

Ex.: Determine the mass in grams of Na+ (22.99 g/mol) in 25.0 g Na2SO4 (142.0

Here the sequence is: GIVEN liters of H2 at STP, CHANGE liters of H2 at STP to

or n = This can also be written: Actual mass (grams) = Number of moles x Molar

(5.8 mol N2O)(44.0124 g N2O/mol N2O) = 260 g N2O. (d) 48.1 . (c) 41.4 g N2H4

n = [ (750.0 mmHg / 760.0 mmHg atm¯1) (0.890 L) ] / (0.08206 L atm mol¯1 K¯1)

Feb 28, 2011 . 7.64 g N2O4 1.56 g NO2. What is the Kc for this reaction? Step 1: Convert grams

Ca(NO3)2(s) + 2 NH4Cl(s) laro.jpg 2 N2O(g) + CaCl2(s) + 4 H2O(g). a 2.00-mol

Jan 13, 2011 . Consider the equilibrium reaction: 2NH3(g) ↔ N2(g) + 3H2(g). . the respective

So the formula for magnesium nitride must be Mg3N2, and we can then

PV = nRT R = 0.0821 atm-L/mol-K. Ideal Gas Example 1. Calculate volume 7.00g

N 2 = 28.0 g O 3 = 48.0 g 1 mole N 2 O 3 = 76.0 g 7.20 mol N 2 O 3 = 547.2 g N 2

4 moles N * (14.0067 g/(mol N)) = 56.0268 g. 2 mol O * (15.9994 g/(mol O)) =

Calculations Using the First Order Rate Equation: r = k[N] . (2.00 mg Co-60)(1 g/

N is equal to molarity or small multiple of molarity N = n M where n = integer .

n= [ mol K ][97.8 KPa ][0.1245dm3] = 0.005 mol gas [8.31 dm3 KPA][18oC + .

1.32 x10–2mol. Do the following problems: 1. Calculate the volume of 36.0 g of

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