Question

11. 13 Consider the following reaction 2 NO, ---> 2N, + 50, The number of moles of o: obtained from 12 moles of N, O, (a) 30 (b) 4.8 (c) 60 (e) none of the above. (d) 6 Consider the reaction 2 N, 50, ---> 2 N, O, If 4 moles of nitrogen is mixed with 96 g of oxygen, what is the limiting reagent? a) , (b) N (c) there is no limiting reagent (d) can't be determined from given data. 12. 13. How many grams of NaOH (molecular mass - 40.0) are needed to prepare 250.0 ml of a 0.500x NaOH solution? (a) 5.00 (b) 5.00x 10 (c) 0.00300 (d) 0.200 (a) none of the above 14. 15. How many milliliters of 8.00 M HCl are required to prepare 2.00 L of a 0.250 M solution of HCI? (a) 0.0625 (b) 16.0 (c) 6.25 (d) 4.00 (a) none of the above How many mls of 0.200 M NaOH are required to react with 20.0 ml of 0.600 MH, SO, the reaction is 2 NaOH + H2So, ---> Na, so + 2 H,0 (a) 30.0 (b) 60.0 (c) 120. (d) 40.0 (a) none of the above 16. If an ideal gas undergoes a change so that its pressure triples, and its volume is cut in hall, the new temperature can be expressed in terms of its original temperatura as (a) I₂ = 3/2 Ti (3) T₂ - 2/3 T, (c) T₂ - T, (@) none of the above. 17. If 40.0 g an ideal gas has a pressure of 2.00 atm and a Volume of 4.00 L and a tenperature of 120. Kthe molecular mass of the gas is (a) 0.20 (b) 32.5 (c) 1.23 (0) 49.3 ( ) none of the above 18. A chamber contains 5 moles of gas A, 4 moles of gas B, and 11 moles of gas C and the total pressure is 5.0 atm. The partial pressure of gas B (in atm) is (a) 1.0 (b) 2.0 (c) 0.10 (d) 4.0 (e) 0.25

in need of everything, take your time

Answer 1

Answer 1- Given,

3.070 * 10⁴

Significant figures = ?

There are 3 significant figures in this

So, OPTION (a) is correct

Answer 2 -Given,

15^{$\degee$$\degree$}C

in ^$\degree$ F and K = ?

We know that,

(0°C × 9/5) + 32 = 32°F

0°C + 273.15 = 273.15 K

So, 15^{$\degee$$\degree$}C = 59°F

15^{$\degee$$\degree$}C = 288.15 K

So, OPTION (a) is correct

Answer 3 - Given,

0.00000328

in scientific notation = ?

0.00000328 = 3.28 * 10^-6

So, OPTION (b) is correct

Answer 4- Given,

Density = 2.78 g/ml

Mass = 34.0 g

Volume in cm³ = ?

We know that,

Density = Mass/ Volume

Volume = Mass/Density

Volume = 34.0 g/2.78 g/ml

Volume = 12.23 ml or 12.23 cm³ [1 ml = 1 cm³]

So, OPTION (a) is correct.

Answer 5 -

6.0 lb/inch²

lb/ft² =?

We know that,

1 lb/inch^{2 =} 144 lb/ft²

So, 6.0 lb/inch^{2 =} 864 lb/ft² or 8.64*10²

So, OPTION (d) is correct.

Answer 6- Given,

13 Ba+

No. of Proton, electron and neutron =?

Chemical symbol for the element Mass number A=Z-N Atomic number = number of protons

So. Number of Protons = 56

Number of Neutrons = 130 -56 = 74

Usually Number of electron are equal to number of protons But it have +2 charge.So, Number of electrons in this is 54-2 = 54

So, OPTION (d) is correct.

Answer 7 -Given,

(NH₄)₂S

%H = ?

In 1 Molecule of (NH₄)₂S there are 8 atoms of H

So, Mass of H = 8 * Molar Mass

Mass of H = 8 * 1 u = 8 u

Molar Mass of (NH₄)₂S = 68.154 g/mol

So, %H = (8u/68.154 g/mol)* 100 = 11.74%

So, OPTION (d) is correct.

Answer 8 -Given,

mass of sample of HF

Number of Molecules of HF = ?

We know that,

Moles = Mass/ Molar Mass

Moles = 60.0 g / 20.01 g/mol = 3

Multiply this by Avogadro's Number (6.022*10²³)

3 mol * 6.022*10²³ = 18.066*10²³ molecules

So, OPTION (b) is correct.

Answer 9 - Given,

N₂ + H₂$\rightarrow$ NH₃

Make the number of atoms of each species equal on both sides

N₂ + 3 H₂$\rightarrow$ 2 NH₃ [Balanced]

Stiochiometric coefficient are 1 of N₂, 3 of H₂, 2 of NH₃

So, OPTION (b) is correct.

Answer 10 - Given,

Mg + O₂$\rightarrow$ NH₃

This is a Combination and redox tyoe reaction.

So, OPTION (d) is correct.

Answer 11 - Given,

2 N₂O₃ $\rightarrow$ 2 N₂ + 5 O₂

The molar ratio of O₂:N₂O₃ = 5:2 i.e 5 moles of O₂ are produced from 2 Moles of N₂O₃

So, moles of O₂ produced = (5/2) * moles of N₂O₃

moles of O₂ produced = (5/2) * 12 = 30

So, OPTION (a) is correct.

Answer 12 -

Given,

Moles of Nitrogen = 4

Mass of Oxygen = 96 g

Moles = Mass/ Molar Mass

Moles of Oxygen = 96 g/32 u = 3 mol

2 N₂ + 5 O₂$\rightarrow$ 2 N₂O₅

For N₂ = 2/4 = 0.5

For O₂ =5/3 = 1.67

So, N₂ is a limiting reagent.

So, OPTION (b) is correct.

Answer 13 -

Given,

Molarity of NaOH = 0.500 M

Volume = 250.0 ml or 0.25 L [1 ml = 0.001 L]

Molar Mass of NaOH = 40 g/mol

Molarity = Mass/(Molar Mass * Volume)

0.500 M = Mass/(40 g/mol * 0.25 L)

Mass = 0.500 M * (40 g/mol * 0.25 L)

Mass = 5 g

So, OPTION (a) is correct.

Answer 14 - Given,

Initial Molarity = 8.00 M

Initial Volume = ?

Final Volume = 2.00 L

Final Molarity = 0.250 M

We know that,

M1* V1 = M2*V2

V1 = (M2*V2/M1)

put the volume,

V1 = (0.250 M * 2.00 L)/8.00M = 0.0625 L or 62.5 ml [1 L = 1000 ml]

So, OPTION (e) is correct.