CO(g) +2H2--->CH3OH
2.50g H2*1mol/2g=1.25 mol H2
30.0L CO*1mol/22.4L=1.34 mol CO,
according to reaction 1 mol CO needs 2 mol H2,so 1.34 mol CO need 2.68 mol H2, so 1) limiting teactant is H2 (H)
2)1.25 mol CH3OH will be produced, 1.25 mol*32g/mol=40.0 g CH3OH
3) 1.25 mol H2 needs 0.625 g CO
1.34-0.625=0.715 g CO leftover
Answer:
[CaSO₄] = 36.26×10⁻² mol/L
Explanation:
Molarity (M) → mol/L → moles of solute in 1L of solution
Let's convert the volume from mL to L
250 mL . 1L/1000 mL = 0.250L
We need to determine the moles of solute. (mass / molar mass)
12.34 g / 136.13 g/mol = 0.0906 mol
M → 0.0906 mol / 0.250L = 36.26×10⁻² mol/L
Answer:
Use the activity formula,
T1/2 = 4.468 x 10^9 yr x 365 x 24 x 3600 = 1.409 x 10^17 sec
l = ln(2)/T1/2 = ln(2)/1.409 x 10^17 = 4.91932697 x 10^-18 s-1
DN/Dt = lN, 265 = 4.91932697 x 10^-18 x N
<u><em>N = 5.38 x 10^19 nuclei</em></u>
Sodium. It reacts with most of every other element.
