Answer:
See Explanation
Explanation:
Given
(a) to (d)
Required
Determine whether the given parameters can calculate the required parameter
To calculate either Density, Mass or Volume, we have
(a) 432 g of table salt occupies 20.0 cm^3 of space
Here, we have:
The above can be used to calculate Density as follows;
(b) 5.00 g of balsa wood, density of balsa wood : 0.16 g/cm^3
Here, we have:
This can be used to solve for Volume as follows:
(c) 32 cm^3 sample of gold density of 19.3 g/cm^3
Here, we have:
This can be used to calculate Mass as follows:
(d) 150 g of iron, density of Iron = 79.0 g/cm^3
Here, we have
This can be used to calculate volume as follows:
<em>Approximated</em>
Answer:
1.37 × 10²³ Atoms of Mercury
Solution:
Step 1: Calculate Mass of Mercury using following formula,
Density = Mass ÷ Volume
Solving for Mass,
Mass = Density × Volume
Putting values,
Mass = 13.55 g.cm⁻³ × 3.4 cm³ ∴ 1 cm³ = 1 cc
Mass = 46.07 g
Step 2: Calculating number of Moles using following formula;
Moles = Mass ÷ M.mass
Putting values,
Moles = 46.07 g ÷ 200.59 g.mol⁻¹
Moles = 0.229 mol
Step 3: Calculating Number of Atoms using following formula;
Number of atoms = Moles × 6.022 ×10²³
Putting value of moles,
Number of Atoms = 0.229 mol × 6.022 × 10²³
Number of Atoms = 1.37 × 10²³ Atoms of Hg
Complete question is;
When a diprotic acid is titrated with a strong base, and the Ka1 and Ka2 are significantly different, then the pH vs. volume plot of the titration will have
a. a pH of 7 at the equivalence point.
b. two equivalence points below 7.
c. no equivalence point.
d. one equivalence point.
e. two distinct equivalence points
Answer:
Option E - Two Distinct Equivalence points
Explanation:
I've attached a sample diprotic acid titration curve.
In diprotic acids, the titration curves assists us to calculate the Ka1 and Ka2 of the acid. Thus, the pH at the half - first equivalence point in the titration will be equal to the pKa1 of the acid while the pH at the half - second equivalence point in a titration is equal to the pKa2 of the acid.
Thus, it is clear that there are two distinct equivalence points.
Answer:
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
Explanation:
Step 1: Data given
iron = Fe = solid = Fe(s)
chlorine = Cl2 = gas = Cl2(g)
iron(III) chloride = FeCl3 = solid = FeCl3(s)
Step 2: The unbalanced equation
Fe(s) + Cl2(g) → FeCl3(s)
Step 3: Balancing the equation
Fe(s) + Cl2(g) → FeCl3(s)
On the left we have 2x Cl (in Cl2) and on the right side we have 3x Cl (in FeCl3). To balance the amount of Cl we have to multiply Cl2 (on the left) by 3 and FeCl3 by 2.
Fe(s) + 3Cl2(g) → 2FeCl3(s)
On the left side we have 1x Fe and on the right side we have 2x Fe (in 2FeCl3). To balance the amount of Fe, we have to multiply Fe on the left side by 2. Now the equation is balanced.
2Fe(s) + 3Cl2(g) → 2FeCl3(s)