Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
Answer:
The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>
Explanation:
Given: The base dissociation constant: 
= 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M
Also, water dissociation constant: 
= 1 × 10⁻¹⁴
<em><u>The acid dissociation constant </u></em>(
)<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>
<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>
Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+
Initial: 0.1 M x x
Change: -x +x +x
Equilibrium: 0.1 - x x x
<u>The acid dissociation constant: </u>![K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5Cleft%20%5BB%20%5Cright%20%5D%20%5Cleft%20%5BH_%7B3%7DO%5E%7B%2B%7D%5Cright%20%5D%7D%7B%5Cleft%20%5BBH%5E%7B%2B%7D%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B%28x%29%28x%29%7D%7B%280.1%20-%20x%29%7D%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%7D%7B0.1%20-%20x%7D)
<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>
Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44
<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>
Answer:
The mass is recorded as 32.075 g
Explanation:
"The first digit of uncertainty is taken as the last significant digit", this is the rule for significant figures in the analysis. The balance measures the mass up to three decimal places, so it makes the most sense to note the whole figure.
It glow, so light energy go out of the system, exotermic
Problem One (left)
This is just a straight mc deltaT question
<em><u>Givens</u></em>
m = 535 grams
c = 0.486 J/gm
tf = 50
ti = 1230
Formula
E = m * c * (ti - tf)
Solution
E = 535 * 0.486 * ( 1230 - 50)
E = 535 * 0.486 * (1180)
E = 301077
Answer: A
Problem Two
This one just requires that you multiply the two numbers together and cut it down to 3 sig digits.
E = H m
H = 2257 J/gram
m = 11.2 grams
E = 2257 * 11.2
E = 25278 to three digits is 25300 Joules. Anyway it is the last one.
Three
D and E are both incorrect for the same reason. The sun and stars don't contain an awful lot of Uranium (1 part of a trillion hydrogen atoms). It's too rare. The other answers can all be eliminated because U 235 is pretty stable in its natural state. It has a high activation complex.
Your best chance would be enriched Uranium (which is another way of saying refined uranium). That would be the right environment. Atomic weapons and nuclear power plants (most) used enriched Uranium. You can google "Little Boy" if you want to know more.
Answer: B
Four
The best way to think about this question is just to get the answer. Answer C.
A: incorrect. Anything sticking together implies a larger and larger result. Gases don't work that way. They move about randomly.
B: Wrong. Heat and Temperature especially depend on movement. Stopping is not permitted. If a substance's molecules stopped, the substance would experience an extremely uncomfortable temperature drop.
C: is correct because the molecules neither stop nor do they stick. The hit and move on.
D: Wrong. An ax splitting something? That is not what happens normally and not with ordinary gases. It takes more energy that mere collisions or normal temperatures would provide to get a gas to split apart.
E: Wrong. Same sort of comment as D. Splitting is not the way these things work. They bounce away as in C.
Five
Half life number 1 would leave 0.5 grams behind.
Half life number 2 would leave 1/2 of 1/2 or 1/4 of the number of grams left.
Answer: 0.25
Answer C
