Answer:
8.33mol/L
Explanation:
First, let us calculate the molar mass of of formaldehyde (CH2O). This is illustrated below:
Molar Mass of CH2O = 12 + (2x1) + 16 = 12 + 2 + 16 = 30g/mol
Mass of CH2O from the question = 0.25g
Number of mole CH2O =?
Number of mole = Mass /Molar Mass
Number of mole of CH2O = 0.25/30 = 8.33x10^-3mole
Now we can calculate the molarity of formaldehyde (CH2O) as follow:
Number of mole of CH2O = 8.33x10^-3mole
Volume = 1mL
Converting 1mL to L, we have:
1000mL = 1L
Therefore 1mL = 1/1000 = 1x10^-3L
Molarity =?
Molarity = mole /Volume
Molarity = 8.33x10^-3mole/1x10^-3L
Molarity = 8.33mol/L
Therefore, the molarity of formaldehyde (CH2O) is 8.33mol/L
The awnser is A. Idek I looked it up so yeah that’s the awnser
Answer:
I think it's C
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Answer:-
Carbon
[He] 2s2 2p2
1s2 2s2 2p2.
potassium
[Ar] 4s1.
1s2 2s2 2p6 3s2 3p6 4s1
Explanation:-
For writing the short form of the electronic configuration we look for the nearest noble gas with atomic number less than the element in question. We subtract the atomic number of that noble gas from the atomic number of the element in question.
The extra electrons we then assign normally starting with using the row after the noble gas ends. We write the name of that noble gas in [brackets] and then write the electronic configuration.
For carbon with Z = 6 the nearest noble gas is Helium. It has the atomic number 2. Subtracting 6 – 2 we get 4 electrons. Helium lies in 1st row. Starting with 2, we get 2s2 2p2.
So the short term electronic configuration is [He] 2s2 2p2
Similarly, for potassium with Z = 19 the nearest noble gas is Argon. It has the atomic number 18. Subtracting 19-18 we get 1 electron. Argon lies in 3rd row. Starting with 4, we get 4s1.
So the short electronic configuration is
[Ar] 4s1.
For long term electronic configuration we must write the electronic configuration of the noble gas as well.
So for Carbon it is 1s2 2s2 2p2.
For potassium it is 1s2 2s2 2p6 3s2 3p6 4s1
Answer:
89.04 g of NaNO₃.
Explanation:
We'll begin by converting 838 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
838 mL = 838 mL × 1 L / 1000 mL
838 mL = 0.838 L
Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:
Volume = 0.838 L
Molarity = 1.25 M
Mole of NaNO₃ =?
Mole = Molarity × volume
Mole of NaNO₃ = 1.25 × 0.838
Mole of NaNO₃ = 1.0475 mole
Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:
Mole of NaNO₃ = 1.0475 mole
Molar mass of NaNO₃ = 23 + 14 + (16×3)
= 23 + 14 + 48
= 85 g/mol
Mass of NaNO₃ =?
Mass = mole × molar mass
Mass of NaNO₃ = 1.0475 × 85
Mass of NaNO₃ = 89.04 g
Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.