Answer:
Do not form hydrogen bonds amongst themselves and have higher vapor pressure.
Explanation:
First, we need to determine the half reaction of magnesium. It would be expressed as:
Mg2+ + 2e- = Mg
Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:
4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C
We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.
35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
<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:
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Explanation:
Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g