Answer:
12.99
Explanation:
<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of KOH: 716. mg (0.716 g)
- Volume of the solution: 130. mL (0.130 L)
Step 2: Calculate the moles corresponding to 0.716 g of KOH
The molar mass of KOH is 56.11 g/mol.
0.716 g × 1 mol/56.11 g = 0.0128 mol
Step 3: Calculate the molar concentration of KOH
[KOH] = 0.0128 mol/0.130 L = 0.0985 M
Step 4: Write the ionization reaction of KOH
KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)
The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M
Step 5: Calculate the pOH
We will use the following expression.
pOH = -log [OH⁻] = -log 0.0985 = 1.01
Step 6: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -1.01 = 12.99
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4p²
Explanation:
This atom will likely have 4 electron shells denotation of – 2.8.8.4
Orbitals shells show the probability, in space around the nucleus, where to find an electron. It is important to note that the 3rd shell has an additional d orbital (-in addition to s and p). However, because the d orbital has a higher energy state than the 4s and 4p orbitals, the d orbital only fills up when these latter ones are completely filled. In this case, the 4p does not completely fill (hence we don't see the d orbital in the notation).
Answer:
Zero
Explanation:
Recall that;
E = q + w
Where;
q = heat, w = work done
When heat is absorbed by the system q is positive
When heat is evolved by the system q is negative
When the system does work, w is negative
When work is done on the system w is positive
Step 1
ΔE1= 60 KJ + 40 KJ = 100KJ
Step 2
ΔE2= (-30 KJ) + (-70 KJ) = (-100) KJ
ΔE1 + ΔE2= 100KJ + (-100) KJ = 0KJ
Answer:
the answer would be A. cells
Explanation:
just trust me Im a 8th grader and I did that before
There are no states in the picture, but Na should have a "(s)" after it, and Cl2 should have a "(g)" after it. NaCl should have an "(s)". Chlorine is a diatomic element so it has a "2" subscript on it.
Hope this helped! :)
