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Answer:
b. 2.28 M
Explanation:
The reaction of neutralization of NaOH with H2SO4 is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
<em>Where 2 moles of NaOH react per mole of H2SO4</em>
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To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:
<em>Moles H2SO4:</em>
45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4
<em>Moles NaOH:</em>
0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH
<em>Molarity NaOH:</em>
0.0457moles NaOH / 0.020L =
2.28M
Right option:
<h3>b. 2.28 M</h3>
When KCL, which is an ionic compound is added to water it will dissociate and or ionize completely forming the ions of K+ and Cl-. The resulting solution would be a neutral solution, as the K+ is a cation of a strong base and Cl- is an anion of a strong acid, and whenever a strong base reacts with a strong acid, a neutral salt is produced
Answer: grams;mass
Explanation: :) I took the test.
Answer:
stay the same.
Explanation: Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):
Na: 1s2 2s2 2p6 3s1
Ar: s2 2s2 2p6 3s2 3p6
As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:
Na: 3s1
Ar: 3s2, 3p6
All the new electrons electrons filled the third energy level (3s and 3p). So the energy level does not change, just the orbitals.
