If I have 15.40mL of 1.4M HBr(aq) and add 22.10mL KOH what is the molarity ofKOH being used?
1 answer:
The molarity of KOH is 0.98 M.
<u>Explanation:</u>
KOH + HBr →KBr + H₂O
As per the above reaction, equal moles of KBr and HBr reacts to form 1 mole of KBr and 1 mole of water. We have to find the molarity of KOH by using the law of Volumetric analysis using the equation as,
V1M1 = V2M2
Here V1 and M1 are the volume and molarity of HBr.
V2 and M2 are the volume and molarity of KOH .
We can find the molarity of KOH as,
Plugin the values as,
= 0.98 M
So the molarity of KOH is 0.98 M.
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5. 25 x 10⁻⁷mg
Explanation:
This is mass conversion from mg to kg;
The kg is a quantity of mass used to measure the amount of matter in a substance.
Given mass = 5.25 x 10⁻¹³kg
The kilo- is a prefix that denotes 10³
therefore;
1000g = 1kilogram
the milli- is a prefix that denotes 10⁻⁻³
1000mg = 1g
Now that we know this, we can convert:
5.25 x 10⁻¹³kg x = 5. 25 x 10⁻¹³ x 10⁶mg
= 5. 25 x 10⁻⁷mg
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Answer:
3.72 mol Hg
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Density = Mass over Volume
Explanation:
<u>Step 1: Define</u>
D = 13.6 g/mL
54.8 mL Hg
<u>Step 2: Identify Conversions</u>
Molar Mass of Hg - 200.59 g/mol
<u>Step 3: Find</u>
13.6 g/mL = x g / 54.8 mL
x = 745.28 g Hg
<u>Step 4: Convert</u>
<u /> = 3.71544 mol Hg
<u>Step 5: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
3.71544 mol Hg ≈ 3.72 mol Hg