Density for Cu: 8.96
g/ml <span>
Density for Ag: 10.5g/ml</span>
First set up an
equation: mass = density * volume <span>
8.96 * x + 10.5* (10.12 - x) = 105.0
The density of the Cu (8.96) times x (the amount of copper)
added to the density of Ag (10.5) times voume of silver (10.12 - the amount of
copper) should equal the total weight.
</span>
Solving for x:<span>
x = 0.818 mL
So
mass of copper = 8.96 * .818 = 7.33
<span>mass percent of copper = 7.33/105 * 100% = 6.98%</span></span>
3 bc Hf is an element hafnium and HF is compound hydrogen flouride
Answer:
The proper matching is given below.
Explanation:
a Separate molecules by size size exclusion chromatography
b Separate molecules by charge Ion exchange chromatography
c The stationary phase has a covalently bound group to which a protein in the mobile phase can bind. Affinity chromatography
d uses mobile phase and stationary phase to separate protein Size exclusion chromatography
e The stationary phase contain cross linked polymers with different pore size
Size exclusion chromatography
f can separate molecules based on protein ligand binding Affinity chromatography
g The stationary phase may contain negatively or positively charged groups
ion exchange chromatography
Answer:
6
Explanation:
Normally, oxygen has 8 electrons and 8 protons. If the charge is 2-, then that means there are two more protons than electrons. There would be 6 electrons.
Answer:
13.5 * 10^-2 g
Explanation:
What we know:
Balanced Equation: 3Ba+Al2(SO4)3 -->2Al+3BaSO4,
Grams of Ba: 1
Grams of Al2(SO4)3: 1.8g
Calculate the # of moles of Ba and Al2(SO4)3:
1g Ba/137.3 = 7.3 *10^-3 mol Ba
1.8g Al2(SO4)3/ 342 = 5.3 *10^-3 mol Al2(SO4)3
Find the limiting reactant:
Ba has a coefficient of 3 in the balanced equation, so we divide the # of moles of Ba by 3 to get... 7.3 *10^-3 mol Ba/3 = 2.43 *10^-3
Al2(SO4)3 has a coefficient of 1, so if we divide by 1, we get the same number of 5.3 *10^-3
2.43 *10^-3 is smaller than 5.3 *10^-3, therefore Ba is the limiting reactant.
finally, we just find the number of moles of Al
The ratio of Al to Ba is 2:3 so...
7.3 * 10^-3 * (2/3) = 5 *10^-3 mol Al
CONVERT TO GRAMS
5 *10^-3 mol Al * 27 = 13.5 * 10^-2 g
<u>Hope that was helpful! </u>