Answer:
1.9×10^20
Explanation:
ϕ = number of reaction events /number of photons absorbed
ϕ= 0.26
Energy= power× time
885×10^-3×10×60= 531J
But E= nhc/λ
n= number of photons
h= planks constant
c= speed of light
λ= wavelength
n= Eλ/hc
n= 531×280×10^-9/6.6×10^-34 ×3×10^8
n= 7.5×10^20
Therefore
From
ϕ = number of reaction events /number of photons absorbed
Number of reaction events= 0.26×7.5×10^20
= 1.95×10^20
Answer: 24.13 g Cu
Explanation:
<u>Given for this question:</u>
M of CuO = 30 g
m of CuO = 79.5 g/mol
Number of moles of CuO = (given mass ÷ molar mass) = (30 ÷ 79.5) mol
= 0.38 mol
The max number of CuO (s) that can be produced by the reaction of excess methane can be solved with this reaction:
CuO(s) + CH4(l) ------> H2O(l) + Cu(s) + CO2(g)
The balanced equation can be obtained by placing coefficients as needed and making sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side
4CuO(s) + CH4(l) ----> 2H2O(l) + 4Cu(s) + CO2(g)
From the stoichiometry of the balanced equation:
4 moles of CuO gives 4 moles of Cu
1 mole of CuO gives 1 mol of Cu
0.38 mol of CuO gives 0.38 mol of Cu
Therefore, the grams of Cu that can be produced = 0.38 × molar mass of Cu
= 0.38 × 63.5 g
= 24.13 grams
Therefore, 24.13 grams of copper could be produced by the reaction of 30.0 of copper oxide with excess methane
68 kg. There are 58 kg salt in 500 gal seawater.
<em>Step 1.</em> Convert gallons to litres
1 US gal = 3.79 L (1 Imp gal = 4.55 L)
<em>Step 2</em>. Find the volume of the seawater
Volume = 500 gal × (3.79 L/1 gal) = 1895 L
<em>Step 3</em>. Find the mass of the seawater
Mass = 1895 L × (1.025 kg/1 L) = 1942 kg
<em>Step 4</em>. Find the mass of the salt
Mass of salt = 1942 kg seawater × (3.5 kg salt/100 kg seawater) = 68 kg salt
