Answer:
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Explanation:
When copper (ii) oxide is heated in presence of hydrogen gas it produces copper metal and water vapor.
that is
CuO + H2 = Cu + H2O
The moles of copper metal produced is calculated as below
find the moles of CUO used
moles = mass/molar mass
=5.6/79.5 =0.07 moles
by use of mole ratio between CUO:Cu which is 1:1 the moles of Cu is also
=0.007 moles of Cu
False it was rutherford if i am not mistaken
<h3>Answer:</h3>
0.8133 mol
<h3>Solution:</h3>
Data Given:
Moles = n = ??
Temperature = T = 25 °C + 273.15 = 298.15 K
Pressure = P = 96.8 kPa = 0.955 atm
Volume = V = 20.0 L
Formula Used:
Let's assume that the Argon gas is acting as an Ideal gas, then according to Ideal Gas Equation,
P V = n R T
where; R = Universal Gas Constant = 0.082057 atm.L.mol⁻¹.K⁻¹
Solving Equation for n,
n = P V / R T
Putting Values,
n = (0.955 atm × 20.0 L) ÷ (0.082057 atm.L.mol⁻¹.K⁻¹ × 298.15 K)
n = 0.8133 mol
Aldehydes and ketones having
α-hydrogen atoms, undergoes aldol condensation, in present of base (NaOH).
The initial product formed during this reaction is
β-hydroxy alcohol, which then undergoes dehydration to form
α,β-unsaturated aldehyde or ketone.
In present case, 3,3-dimethyl-2-butanone has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
5-hydroxy-4,4-dimethly-1-(2-methylcyclopentyl)pent-1-en-3-one. The complete reaction product formed is shown below.