Answer:
Both oil and gasoline molecules are nonpolar, while water is polar. Nonpolar solvents have a tendency to dissolve other nonpolar molecules.
Explanation:
Molecules may be categorized as "polar" or "nonpolar" according to <em>difference in the atom's electronegativity.</em>
<u>Water is polar</u> because it consists of two types of atoms that<em> do not cancel out each other.</em> It is made of two atoms of Hydrogen and only one atom of Oxygen. This makes the Oxygen<u> partially negative</u> and the Hydrogen <u>partially positive.</u> This allows them to readily bond with other polar molecules like sugar. However, it cannot mix freely with oil and gasoline because<em> both of these are nonpolar. </em>Nonpolar molecules do not have much difference when it comes to their atoms' electronegativity. <em>Therefore, they have the tendency to dissolve molecules which are nonpolar as well. </em>This explains why oil molecules can mix freely with gasoline.
First write out the balanced equation. 3Cu+2Ag(NO3)3=2Ag+3Cu(NO3)2
Then convert copper from grams to moles
15 g*1 mol cu/63.54 g= 15/63.54 mol cu
Then use the mole ratio to convert Moles Cu to Moles Ag
15/63.54 moles Cu* 2 moles Ag/3 moles Cu
The final awnser is (15*2)/(63.54*3) moles Ag =0.157 moles Ag. If the question wants the answer in grams, convert from moles Ag to grams Ag.
0.157 moles Ag*107.87 g Ag/ mol Ag=16.98 g Ag
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
Answer:
A = 349 g.
Explanation:
Hello there!
In this case, since the radioactive decay kinetic model is based on the first-order kinetics whose integrated rate law is:
We can firstly calculate the rate constant given the half-life as shown below:
Therefore, we can next plug in the rate constant, elapsed time and initial mass of the radioactive to obtain:
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