Answer: it’s B!! :)
C. If an electron is added to the outer shell, how will this affect the charge and volume of the atom?
if an electron is added on the outer shell
Answer:
An atom that gains or loses an electron becomes an ion. If it gains a negative electron, it becomes a negative ion. If it loses an electron it becomes a positive ion.
Answer:
when an atom gains an electron, it gains a negative charge . it's volume increases as well due to the increase in repulsion between the electrons
Which of the following is NOT a
homogenous mixture?
Answer:blacks screen. No answers to choose from. Thanks for the points:))
Explanation:
lab
table b:eyedrop of blue water in cold water
table c:eyedrop of red water in hot water
table d:see how temperature changes on white + black paper with light shining on them
plz help
brainliest for 1st correct answer
Answer:
lol, I just did this assignment, cold water
0:30: slowly expand
1:00: still expanding
1:30: expanding is continued
2:00: still expanding slowly
2:30: expanding slowly
3:00: expanded twice the size from the first time
hot water
0:30: expanding quickly
1:00: still expanding
1:30: starts to reach the surface and still expanding
2:00: almost all the of the dye reached the top
2:30: the dye reached the top and is starting to split
3:00: the dye split appart and is moving down again
radiation
white paper
0:30: at a low temp. at around 3.3 C
1:00: still at a low temp. around 4.4 C
1:30: slowly rising temp. around 5.5 C
2:00: still slowly rising temp. around 7.2 C
2:30: still at a low temp. around 8.3 C
3:00: ends at a low temp. around 10 C
black paper
0:30: at a resonable temp. at around 6.6 C
1:00: temp. rises quicker at around 13.3 C
1:30: temp. stil rising quickly at around 17.7 C
2:00: temp. is getting hotter at around 21.6 C
2:30: temp. is still getting hotter at around 26.1 C
3:00: temp. ends at a hot temp. at around 30 C
Explanation:
What is the pH after two solutions are mixed: 1. A 0.155 M H2SO4 solution with a volume of 35.0 mL and 2. A 0.225 M solution of KOH with a volume of 35.0 mL. The volumes are additive.
Answer:
[tex]pH=1.67[/tex]
Explanation:
Hello.
In this case, since the reaction between the sulfuric acid and the potassium hydroxide is:
[tex]2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O[/tex]
We can see a 2:1 mole ratio between them, thus, we compute the moles of acid and base that are present in the solution by using the employed volume and its concentration:
[tex]n_{H_2SO_4}=0.155mol/L*0.0350L=0.005425molH_2SO_4\\\\n_{KOH}=0.225mol/L*0.0350L=0.007875molKOH\\[/tex]
Next, we use the 2:1 mole ratio to compute the consumed moles of acid:
[tex]n_{H_2SO_4}^{consumed}=0.007875molKOH*\frac{1molH_2SO_4}{2molKOH}=0.0039375molH_2SO_4[/tex]
It means that the leftover of sulfuric acid is:
[tex]n_{H_2SO_4}^{leftover}=0.005425-0.0039375=0.001488molH_2SO_4[/tex]
Therefore, since sulfuric acid is a strong acid, the pH is defined in terms of its concentration which is actually equal to the concentration of hydrogen ions in solution, which is computed considering the total additive volume:
[tex][H^+]=[H_2SO_4]=\frac{0.001488mol}{0.070L}=0.02125M[/tex]
So the pH turns out:
[tex]pH=-log(0.02125)\\\\pH=1.67[/tex]
Best regards.