Although the process varies slightly from one material to another, the general process is as follows:
1) Choose an appropriate container for the solid. This may be a petri dish or a beaker in which you want to prepare the solution of the solid or any other lab equipment.
2) Place the container on a mass balance, then turn the balance on. The mass balance will automatically zero-out the mass of the container, so that any mass that you add on the container will be the mass of the solid. Alternatively, you may first measure the mass of the empty container alone.
3) Add the solid using a lab spatula. The solid should be added more slowly when the reading on the scale comes close to the desired value.
4) Remove the container from the mass balance after the desired amount of solid has been added.
Complete question:
Consider the reaction.
At equilibrium at 600 K, the concentrations are as follows.
2HF -----> H₂ + F₂
[HF] = 5.82 x 10-2 M
[H2] = 8.4 x 10-3 M
[F2] = 8.4 x 10-3 M
What is the value of Keq for the reaction expressed in scientific notation?
2.1 x 10-2
2.1 x 102
1.2 x 103
1.2 x 10-3
Answer:
2.1 × 10^-2
Explanation:
Kequilibrum(Keq) = product/reactant
Equation for the reaction :
2HF -----> H₂ + F₂
Therefore,
Keq = [H2][F2] / [HF]^2
Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2
Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]
Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]
Keq = 2.0665 × 10^(-6 - (-4))
Keq = 2.0665 × 10^(-6 + 4)
Keq = 2.1 × 10^-2
The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar
Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691
CaCO3(s) ⟶ CaO(s)+CO2(s)
<span>
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 </span>
<span>
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3,
therefore 0.0131 moles CO2 should also be formed.
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 </span>
Therefore:<span>
<span>% Yield: 0.53/.576 x100= 92 percent yield</span></span>
