Answer:
The answer to your question is 2.32 atm
Explanation:
Data
P = ?
n = 0.214
V = 2.53 L
T = 61°C
R = 0.082 atm L/mol°K
Formula
PV = nTR
solve for P
P = nRT/V
Process
1.- Calculate the temperature in K
°K = °C + 273
°K = 61 + 273
= 334
2.- Substitution
P = (0.214 x 0.082 x 334) / 2.53
3.- Simplification
P = 5.86/2.53
4.- Result
P = 2.32 atm
Answer:
Explanation:
A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters
PV =nRT
at STP P= 1atm. T= 273 K
n is the number of moles. O2 has a molar mass of 32.
30 gm of O2 is 30/32= 0.94 =n
PV = nRT
at STP: P= 1 atm, T=273 K, R is always 0.082 for P in atm and T in K
SO
1 X V = 0.94 X 0.082 X 273
using high school freshman algebra,
V= 0.94 X 0.082 X 273 = 21L
using high school algebra I,
V=
Well they buy food then they cook it then they eat it
Answer:
1). 1 mole of Carbon burnt in air
C + O2 →CO2
1 mole of carbon produces 1 mole of CO2 which is 44g of CO2
2). 1 mole of carbon is burnt in 16g of dioxygen
32g of O2 = 44g of CO2
1g of O2 = 44/32
CO2 (Dioxygen is limiting reagent)
16g of O2 = 4/32 × 16 = 22g of CO2 in one mole
3) 2 moles of Carbon burnt in 16g of dioxygen
16g of dioxygen is available, and thus it can combine with 0.5 mol of carbon to give 22g of CO2
