BaO, Barium Oxide.
Na2SO4, Sodium Sulfate.
CuO, Copper (II) Oxide.
P2O5, Diphosphorus Pentoxide.
HNO3, Nitric Acid.
CO32-, Molecular Formula.
Hope this helps. :)
Answer:
D. Scientific notation is a way of writing large and small numbers
Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
Answer:
n l m
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1 0 0 1s 1 2 2
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2 0 0 2s 1 2
2 1 1,0,-1 2p 3 6 8
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3 0 0 3s 1 2
3 1 1,0,-1 3p 3 6
3 2 2,1,0,-1,-2 3d 5 10 18
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4 0 0 4s 1 2
4 1 1,0,-1 4p 3 6
4 2 2,1,0,-1,-2 4d 5 10
4 3 3,2,1,0,-1,-2,-3 4f 7 14 32
Explanation:
