The answer is <span>D.when the aim is to show electron distributions in shells. This is because there are some instances when elements don't possess a regular or normal electron configuration. There are those who have special electron configurations wherein a lower subshell isn't completely filled before occupying a higher subshell. It is best to visualize such cases using the orbital notation.</span>
Answer:
7.46 g
Explanation:
From the balanced equation, 2 moles of Mg is required for 2 moles of MgO.
The mole ratio is 1:1
mole = mass/molar mass
mole of 4.50 g Mg = 4.50/24.3 = 0.185 mole
0.185 mole Mg will tiled 0.185 MgO
Hence, theoretical yield of MgO in g
mass = mole x molar mass
0.185 x 40.3 = 7.46 g
Use ideal gas equation: pV = nRT
Now pass n to mass: n = mass / MM .... [MM is the molar mass]
pV = [mass/MM]*RT =>mass/V = [p*MM] / RT and mass / V = density
p= 130 kPa = 130,000 Pa = 130,00 joule / m^3
T = 10.0 ° + 273.15 = 283.15 k
MM of sulfur (S) = 32 g/mol = 32000 kg/mol
density = 130,000 Pa * 32000kg/mol / [8.31 joule / mol*k * 283.15 k] = 1.77*10^6 kg/m^3 = 1.77 g/L ≈ 1.8 g/L
Then, I do not get any of the option choices.
Is it possbile that the pressure is 13.0 kPa instead 130. kPa? If so the answer would be 18 g/L
Note that the mass is not used. You do not need it unless you are asked for the volume, which is not the case.
<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
Answer:
Explanation:
For a chemical reaction, the enthalpy of reaction (ΔHrxn) is … ... to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g•°C). ... Both Equations 12.3.7 and 12.3.8 are under constant pressure (which ... The specific heat of water is 4.184 J/g °C (Table 12.3.1), so to heat 1 g of water by 1 ..
