Answer:
Original temperature (T1) = - 37.16°C
Explanation:
Given:
Gas pressure (P1) = 2.75 bar
Temperature (T2) = - 20°C
Gas pressure (P2) = 1.48 bar
Find:
Original temperature (T1)
Computation:
Using Gay-Lussac's Law
⇒ P1 / T1 = P2 / T2
⇒ 2.75 / T1 = 1.48 / (-20)
⇒ T1 = (2.75)(-20) / 1.48
⇒ T1 = -55 / 1.48
⇒ T1 = - 37.16°C
Original temperature (T1) = - 37.16°C
<span>the answer is 0.2 ATM</span>
Answer:
13.2 g Na2CO3
Explanation:
Convert 10.0 g NaOH to mol.
10.0 g x 1 mol/39.997 g = 0.250 mol
Use mol ration given by the equation: 2 mol NaOH to 1 mol Na2CO3
0.250 mol NaOH x 1 mol Na2CO3/2 mol NaOH = 0.125 mol Na2CO3
Finally, convert the moles of Na2CO3 to grams.
0.125 mol Na2CO3 x 105.99 g/1 mol = 13.2 g
