Answer:
Explanation:
n CaCO3 = mass / m.wt
= 500 /( 40 + 12 + 16x 3)
= 5 mole
n CaO = 5 moles ( from the balanced equation we have 1:1 moles )
mass of CaO = nCaO X m.wt
5 x( 40 +16 )
= 280 grams
Answer:
V = 85.2
Explanation:
STP = 273K and 1 atm
Considering what we know about STP, we get the moles, temperature, and pressure. Using the ideal gas law we can find the volume (PV = nRT). Plug in our variables: (1 * V = 3.80 * R * 273). Since we are dealing with atm and not kPA or mmHg, we use the constant for atm (0.0821) which we use for R. (So.. now our equation is 1 * V = 3.80 * 0.0821 * 273). We now multiply the right side to get 85.17054. So... V = 85.2 considering sigificant figures (this is the part where I am the least sure of, since I havent done sig figs in a while)
This is what I got. Hope it helps :)
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
<span>They want a full outer shell of electrons, so the lose, gain, or share electrons with other elements, forming compounds, until they have 8 valence electrons and become stable.
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