A 25 ml sample of gas is in a syringe at 22C if it was cool down to 0 C what will the volume of the gas become

0.023 liters

Explanation:

Given that,

Original volume of gas (V1) = 25mL

[convert 25mL to liters

If 1000ml = 1L

25ml = 25/1000 = 0.025L]

Original temperature of gas (T1) = 22°C

[Convert 22°C to Kelvin by adding 273

22°C + 273 = 295K]

New volume of gas (V2) = ?

New temperature of gas (T2) = 0°C

[Convert 0°C to Kelvin by adding 273

0°C + 273 = 273K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

0.025L/295K = V2/273K

To get the value of V2, cross multiply

0.025L x 273K = 295K x V2

6.825L•K = 295K•V2

Divide both sides by 295K

6.825L•K/295K = 295K•V2/295K

0.023 L = V2

Thus, the new volume of the gas will be 0.023 litres

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g