Answer:
0.51M
Explanation:
Given parameters:
Initial volume of NaBr = 340mL
Initial molarity = 1.5M
Final volume = 1000mL
Unknown:
Final molarity = ?
Solution;
This is a dilution problem whereas the concentration of a compound changes from one to another.
In this kind of problem, we must establish that the number of moles still remains the same.
number of moles initially before diluting = number of moles after dilution
Number of moles = Molarity x volume
Let us find the number of moles;
Number of moles = initial volume x initial molarity
Convert mL to dm³;
1000mL = 1dm³
340mL gives 
= 0.34dm³
Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles
Now to find the new molarity/concentration;
Final molarity = 
= 
= 0.51M
We can see a massive drop in molarity this is due to dilution of the initial concentration.
Answer:
8
Explanation:
The atomic number refers to the number which identifies the element, which is the proton number.
Answer:
0.282 M
General Formulas and Concepts:
<u>Chemistry - Solutions</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Molarity = moles of solute / liters of solution
Explanation:
<u>Step 1: Define</u>
5.85 g KI
0.125 L
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of I - 126.90 g/mol
Molar Mass of KI - 39.10 + 126.90 = 166 g/mol
<u>Step 3: Convert</u>
<u />
= 0.035241 mol KI
<u>Step 4: Find Molarity</u>
M = 0.035241 mol KI / 0.125 L
M = 0.281928
<u>Step 5: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.281928 M ≈ 0.282 M
I believe it might be A. Damaged buildings.
B is a positive and so is Tourism.
Answer:
pH → 7.46
Explanation:
We begin with the autoionization of water. This equilibrium reaction is:
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴ at 25°C
Kw = [H₃O⁺] . [OH⁻]
We do not consider [H₂O] in the expression for the constant.
[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ M
Kw depends on the temperature
0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻] → [H₃O⁺] = [OH⁻] at 0°C
√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M
- log [H₃O⁺] = pH
pH = - log 3.46×10⁻⁸ → 7.46
