I believe the best answer to that question wud be D. I cud b wrong
ITS A SIMPLE.............................
FuFu Boys
> 2,000
mL of a 5.0 × 10–5% (w/v) sucrose solution
5.0 × 10–3
g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol
<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>
5 grams /
1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol
<span>
> 20 mL of a 5.0 M sucrose solution </span>
5.0 M *
0.020 L = 0.1 mol
Answer:
<span>2,000 mL
of a 5.0 ppm sucrose solution</span>
Answer:
The reaction will move to the left.
Explanation:
<em>Ba(OH)₂ = Ba²⁺ + 2OH⁻,</em>
<em>Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.</em>
- If H⁺ ions are added to the equilibrium:
H⁺ will combine with OH⁻ to form water.
<em>So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.</em>
<em />
<em>According to Le Châtelier's principle: </em>when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
- So, the reaction will move to the right to suppress the effect of decreasing OH⁻ concentration.
- The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.
<em>So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.</em>
Answer:
If the volume of a gas increased from 2 to 6 L while the temperature was held constant, <u><em>the pressure of the gas decreased by a factor of 3.</em></u>
Explanation:
Boyle's law that says "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or
P * V = k
To obtain the proportionality factor k you must make the quotient:

k= 3
This means that <u><em>if the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.</em></u>