Answer:
pH = 11.9
Explanation:
First, we <u>determine the number of OH⁻ moles dissolved</u>:
<em>80% of Ca(OH)₂ is dissolved</em>:
- 0.0005 mol * 80/100 = 4x10⁻⁴ mol Ca(OH)₂
<em>There are two OH⁻ moles per Ca(OH)₂ mol</em>:
- 4x10⁻⁴ * 2 = 8x10⁻⁴ mol OH⁻
Now we can <u>calculate the molar concentration of OH⁻</u> (moles/L):
- 100 mL ⇒ 100/1000 = 0.1 L
- [OH⁻] = 8x10⁻⁴ mol / 0.1 L = 8x10⁻³ M
Then we <u>calculate the pOH of the solution</u>:
- pOH = -log[OH⁻] = -log(8x10⁻³ M) = 2.10
Finally, we can <u>calculate the pH of the solution</u> using the equation
Answer:
see explanation
Explanation:
To determine limiting reactant divide mole quantities of reactants by the respective coefficient in the balanced equation. The smaller value is the limiting reactant.
P₄ + 5O₂ => 2P₂O₅
12/1 = 12 15/5 = 3
O₂ is the limiting reactant. P₄ will be in excess when rxn stops.
In this problem, we can formulate two equations.
First, the overall mass balance of the ointment. The mass of
30% salicylic acid ointment plus the mass of petrolatum must be 50 grams.
x + p = 50
Where x is the mass of 30% salicylic acid ointment and p is
the mass of petrolatum
Second, the component mass balance of salicylic acid:
0.3 * x = 50 * 0.05
x = 8.33 g
Therefore the mass of petrolatum to be used is:
x + p = 50
p = 50 – x
p = 50 – 8.33
p = 41.67 g
Answer:
41.7
Answer:
The correct option is: B. 13g
Explanation:
Given: Molar mass of iron (II) sulfate: m = 260g/mol,
Molarity of iron (II) sulfate solution: M = 0.1 M,
Volume of iron (II) sulfate solution: V = 500 mL = 500 × 10⁻³ = 0.5 L (∵ 1L = 1000mL)
Mass of iron (II) sulfate taken: w = ? g
<em>Molarity</em>:
Here, n- total number of moles of solute, w - given mass of solute, m- molar mass of solute, V- total volume of solution in L
∴ <em>Molarity of iron (II) sulfate solution:</em>
⇒
⇒
⇒ <em>mass of iron (II) sulfate taken:</em>
<u>Therefore, the mass of iron (II) sulfate taken for preparing the given solution is 13 g.</u>
Answer:
proton: positive
electron: negative
neutron: neutral
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