(i) We start by calculating the mass of sugar in the solution:
mass of sugar = concentration × solution mass
mass of sugar = 2.5/100 × 500 = 12.5 g
Then now we can calculate the amount of water:
solution mass = mass of sugar + mass of water
mass of water = solution mass - mass of sugar
mass of water = 500 - 12.5 = 487.5 g
(ii) We use the following reasoning:
If 500 g solution contains 12.5 g sugar
Then X g solution contains 75 g sugar
X=(500×75)/12.5 = 3000 g solution
Now to get the amount of solution in liters we use density (we assume that is equal to 1):
Density = mass / volume
Volume = mass / density
Volume = 3000 / 1 = 3000 liters of sugar solution
Sorry I honestly have no idea I just want my points
Answer:
SO SORRY IF THIS IS WRONG BUT I HOPE THIS HELPS
Explanation:
Primary fertilizers include substances derived from nitrogen, phosphorus, and potassium. Various raw materials are used to produce these compounds. ... The phosphorus component is made using sulfur, coal, and phosphate rock. The potassium source comes from potassium chloride, a primary component of potash.
Answer:
pH = 6.999
The solution is acidic.
Explanation:
HBr is a strong acid, a very strong one.
In water, this acid is totally dissociated.
HBr + H₂O → H₃O⁺ + Br⁻
We can think pH, as - log 7.75×10⁻¹² but this is 11.1
acid pH can't never be higher than 7.
We apply the charge balance:
[H⁺] = [Br⁻] + [OH⁻]
All the protons come from the bromide and the OH⁻ that come from water.
We can also think [OH⁻] = Kw / [H⁺] so:
[H⁺] = [Br⁻] + Kw / [H⁺]
Now, our unknown is [H⁺]
[H⁺] = 7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]
[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) / [H⁺]
This is quadratic equation: [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴
a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴
(-b +- √(b² - 4ac) / (2a)
[H⁺] = 1.000038751×10⁻⁷
- log [H⁺] = pH → 6.999
A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.
c. Isoleucine has a carbon “branched” closer to the alpha carbon than does leucine.
The structure of leucine is CH3CH(<u>CH3</u>)CH2CH(NH2)COOH.
The structure of isoleucine is CH3CH2CH(<u>CH3</u>)CH(NH2)COOH.
In leucine, the CH3 group is <em>two carbons away</em> <em>from</em> the α carbon; in isoleucine, the CH3 group is on the carbon <em>next to</em> the α carbon.
Thus, <em>isoleucine</em> has the closer branched carbon.
“One is charged, the other is not” is i<em>ncorrect</em>. Both compounds are uncharged.
“One has more H-bond acceptors than the other” is <em>incorrect</em>. Each acid has two H-bond acceptors — the N in the amino and the O in the carbonyl group.
“They have different numbers of carbon atoms” is <em>incorrec</em>t. They each contain six carbon atoms.
