First, by magnet separate paper clips.
Then, By handpicking, separate the pebbles.
Then, using a suitable sieve, separate toothpicks and toothpicks.
Through 100 ml of a one-molar sodium hydroxide solution, 896 ml (n.a.) of carbon dioxide was passed gas. Determine the mass solution of salt in the resulting solution. Density of a one-molar alkali solution, take equal to 1.04 g / ml.
Answer:
0.365 m
Explanation:
The <em>definition of molality</em> is:
- molality = moles of solute / kg of solvent
First <u>we calculate the moles of the solute, LiCl</u>. We do so using its <em>molar mass</em>:
- 0.317 g ÷ 42.394 g/mol = 7.48x10⁻³ mol
Then we calculate the mass of the solvent, water. We do so using its <em>density</em>:
- 20.5 mL * 0.9982 g/mL = 20.5 g
- 20.5 g / 1000 = 0.0205 kg
Finally we <u>calculate the molality of the solution</u>:
- 7.48x10⁻³ mol / 0.0205 kg = 0.365 m
Answer:
Rate = k [X]⁻¹ [Z]²
Explanation:
[X] [Y] [Z] initial rate M M M M · s −1
Exp 1 0.30 0.20 0.35 0.210
Exp 2 0.60 0.10 0.70 0.420
Exp 3 0.60 0.20 0.70 0.420
Exp 4 0.60 0.40 0.35 0.105
In Experiment 2 and 3 where the concentrations of Y and Z were constant, doubling the concentration of Y had no effect on the rate of the reaction. This means, that the rate of the reaction is zero order with respect to Y.
In experiment 3 and 4, dividing the concentration of Z by 2, causes the rate of the reaction to decrease by 4. This means the rate of the reaction is second order with respect to Z.
In experiment 1 and 4, doubling the concentration of X, causes the rate of the reaction to decrease by half. This means that X has an order of -1 with respect to the rate of the reaction.
The rate expression is given as;
Rate = k [X]⁻¹[Y]⁰[Z]²
Rate = k [X]⁻¹ [Z]²
