Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
Answer:
0.9975 cup
Step-by-step explanation:
"Unrefined dark crystalline sugar" is what non-chemists call "brown sugar."
200.0 g brown sugar = 1 cup
199.5 g brown sugar = 199.5× 1/200
.0
199.5 g brown sugar = 0.9975 cup
A standard measuring cup is not capable of this precision and, furthermore, the mass of brown sugar you can get into a cup depends on how tightly you pack it.
Your Mole Day cake will be fine if you use 1 cup of brown sugar as usual.
Answer:
I attached a photo of balanced equations but thats as much as I can help.
Explanation:
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
Answer:

Explanation:
1. Calculate the initial moles of acid and base

2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)