If the [OH-] of a solution is 2.7 x 10-4 mol/L the pOH of the solution is
Answer:
14/2.1 thats just what i do
Explanation:
If the [OH⁻] of a solution is 2.7 × 10⁻⁴ mol/L the pOH of the solution is 3.569.
What is pH ?pH is used to measure whether the substance is acidic, basic or neutral and the range is 0-14.
How to find the pOH of the solution ?To calculate the pOH it is expressed as:
pOH = - log[OH⁻]
= - log[2.7 × 10⁻⁴]
= - [log 2.7 + log 10⁻⁴] [log a × b = log + log b]
= - [log 2.7 - 4 log 10] [log aˣ = x log a]
= - [log 2.7 - 4 × 1] [log 10 = 1]
= - [log 2.7 - 4]
= - [0.431 - 4]
= - [-3.569]
= 3.569
Thus, from above conclusion we can say that If the [OH⁻] of a solution is 2.7 × 10⁻⁴ mol/L the pOH of the solution is 3.569.
Learn more about the pH here: https://brainly.com/question/24595796
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for an electron to move from an energy level that was close to the nucleus to an energy level that is far from the nucleus would it need to gain or lose energy ?
Answer:
it would need to lose energy
Explanation:
I don't really have an explanation
As the electrons gets closer to the nucleus, by the electrostatic force of attraction electrons are tightly bound to the nucleus. To move the inner electron to outer shell, high energy have to applied to weaken the nuclear force.
What is nuclear force?Nuclear force is the force by which the nucleon are bind in the nucleus and electrons are attracted towards the nucleus. The positively charged protons inside the nucleus electrostatically attracts the negatively charged electrons.
The minimum amount of energy required to remove an electron from the valence shell by overcoming the nuclear force is called ionization energy.
To bring an electron which is closer to the nucleus to the outer shell needs much energy to reduce the nuclear attractive pull. The energy need to gain.
Find more on ionization energy:
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I need help working these
Answer:
1) 7.80 X 10^22[tex]\frac{1}{6.02*10^{23} }[/tex] = .13
2) 5.6 [tex]\frac{6.02*10^{23} }{1}[/tex] = 3.4 X 10^24
3)3.81 [tex]\frac{1}{74.55}[/tex] X [tex]\frac{6.02*10^{23} }{1}[/tex] = 3.08 X 10^22
4) 399.91 [tex]\frac{4.05*10^{23} }{6.02*10^{23} }[/tex] = 269
5)4.32 [tex]\frac{1}{18.01}[/tex] X [tex]\frac{22.4}{1}[/tex] = 5.37
Explanation:
For 2 the correct answer is 3.4 instead of 3.2. The 3.4 is rounded from 3.37. Sadly I couldn't figure out the bonus.
PLS HELP THIS IS DUE TODAY!!
Answer:
I think it would be Yellow
PLEASE ANSWER, I WILL GIVE BRAINLIEST AND THANKS!
48g of magnesium and 32g of oxygen react to form 80g of magnesium oxide. 2Mg + O2 --> 2MgO What mass of oxygen is needed to make 20g of MgO?
Answer:
8g
Explanation:
1mole of oxygen -->2moles MgO
32g --> 80g
?g --> 20g
(20x32)÷80
= 8g oxygen
Answer:
Mass of oxygen needed = 8 g
Explanation:
Given data:
Mass of oxygen needed = ?
Mass of MgO formed = 20 g
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of MgO:
Number of moles = mass/molar mass
Number of moles = 20 g/ 40 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of MgO and O₂:
MgO : O₂
2 : 1
0.5 : 1/2×0.5 = 0.25 mol
Mass of oxygen needed:
Mass = number of moles × molar mass
Mass = 0.25 mol × 32 g/mol
Mass = 8 g