Answer:
2C3H8O + 9O2 ==> 6CO2 + 8H2O ... balanced equation
moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles
moles O2 = 31.8 g x 1 mol/31.9 g = 0.997 moles O2
Propanol is limiting based on the mol ratio in balance equation of 2 : 9
To find mass of O2 (excess reagent) left over, we will first find moles O2 used up.
moles O2 used = 0.0875 mol propanol x 9 mol O2/2 mol propanol = 0.394 moles O2 used
moles O2 left over = 0.997 mol - 0.394 mol = 0.603 mol O2 left
mass O2 left = 0.603 mol O2 x 32 g/mol = 19.3 g O2 left over
1. Julie had saved 105 coins from her daily allowance. If the coins consist of 5-peso coins and 10-peso coins amounting to ₱950, how many of each kind of coin did she have?
1. Julie had saved 105 coins from her daily allowance. If the coins consist of 5-peso coins and 10-peso coins amounting to ₱950, how many of each kind of coin did she have?
You did not provide possible answers, but one possible might be that the current atomic theory is so sound and plausible that there should not be anything that could change it in the near future.
Answer:
1.84 L
Explanation:
Using the equation for reversible work:
Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( - 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
