Change in enthalpy when 11.2 dm3 of helium at ntp is heated in a cylinder to 100 degrees celsius

Enthalpy change = 333.3 J

Explanation:

Given data:

Volume of helium = 11.2 dm³ or 11.2 L

Temperature = 100 °C (100+273= 373 K)

Enthalpy change = ?

Solution:

Normal temperature of room = 20°C (20+273 = 293 K)

one mole of gas occupy volume = 22.4 L

ΔT = final temperature - initial temperature

ΔT = 373K - 293K

ΔT = 80K

Number of moles of gas = 11.2 L/22.4 l

Number of moles of gas = 0.5 mol

Enthalpy change = nRΔT

Enthalpy change = 0.5 mol × 8.3j/mol.k ×80 K

Enthalpy change = 333.3 J

linda: 3 ; laura: 15

step-by-step explanation:

note that linda + laura's age is 18, and that linda is 12 years younger than her sister.

let "linda" = x , "laura" = y:

x + y = 18

x = y - 12

first, plug in y - 12 for x in the first equation:

(y - 12) + y = 18

simplify.

2y - 12 = 18

isolate the variable (y). note the equal sign, what you do to one side, you do to the other. do the opposite of pemdas. first add, then divide.

2y - 12 (+12) = 18 (+12)

2y = 18 + 12

2y = 30

(2y)/2 = (30)/2

y = 30/2

y = 15

next, plug in 15 for y in the second equation (the other equation would work too).

x = y - 12

x = (15) - 12

x = 3

linda's age is 3, laura's age is 15.

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