Balance Chemical Equation for this reaction is,
2 CH₄ + O₂ → 2CH₃OH
According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,
22.4 L O₂ required to consume = 44.8 L CH₄
0.660 L O₂ will require = X L of CH₄
Solving for X,
X = (44.8 L × 0.660 L) ÷ 22.4 L
X = 1.320 L of CH₄
Result:
1.320 L of CH₄ <span>gas is needed to react completely with 0.660 L of O</span>₂<span> gas to form methanol (CH</span>₃OH<span>).</span>
Answer:
Where Blocal = local magnetic field between the two regions of the molecule
Blocal = (1-σ)B0
ΔBlocal = (1-σ1)B0 - (1-σ2)B0 = (σ2 - σ1)B0 = ΔσB0 ≈ ΔδB0 x 10∧-6
= (3.36-1.16) x 10∧-6 x B0 = 2.20 x 10∧-6B0
(a) ΔBlocal = 2.20 x 10∧-6 x 1.9T = 4.2 μT
(b) ΔBlocal = 2.20 x 10∧-6 x 16.5T = 36.3 μT
Explanation:
Answer:
Explanation:
The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.
The mass of hydrated salt - 2.123 g
mass of anhydrous salt - 1.861 g
mass that has been reduced is the mass of water that has been heated and lost from the compound thereby making the salt anhydrous.
therefore mass of water lost - 2.123 - 1.861 = 0.262 g
number of moles of water lost - 0.262 g / 18 g/mol = 0.0146 mol
number of moles of salt - 1.861 g / 380.6 g/mol = 0.00490 mol
molar ratio of moles of water to moles of salt
molar ratio = 0.146 mol / 0.00490 mol = 2.98 rounded off to 3
for every 1 mol of salt there are 3 moles of water
therefore empirical formula - Cu₃(PO₄)₂.3H₂O
