What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa

A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Explanation:

Given: = 237 K,   = 101.33 kPa,       = ?

= 300 K,       = 151.99 kPa,         = 0.50 L

Formula used is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

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

you did not give any options thus this is the best answer i can give to you with this little information you have given

> they are both natural and not man made

> they can be tracked and predicted by the right tools

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