An arctic weather balloon is filled with 46.0L of helium gas inside a prep shed. The temperature inside the shed is 12°C The balloon is then taken outside, where the temperature is -10°C Calculate the new volume of the balloon. You may assume the pressure on the balloon stays constant at exactly 1atm Be sure your answer has the correct number of significant digits.
please help

42.45 L

Explanation:

Using the Charle's law equation as follows:

V1/T1 = V2/T2

Where;

T1 = initial temperature (K)

T2 = final temperature (K)

V1 = initial volume (L)

V2 = final volume (L)

Based on the information provided in this question;

V1 = 46.0 L

V2 = ?

T1 = 12°C = 12 + 273 = 285K

T2 = -10°C = -10 + 273 = 263K

Using V1/T1 = V2/T2

46/285 = V2/263

Cross multiply

285V2 = 46 × 263

285V2 = 12098

V2 = 12098 ÷ 285

V2 = 42.45L

The new volume of the balloon is 42.45 L.

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

the ratio of the conjugate base to the acid required to attain the desired ph.