Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
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Answer:
wheres the question or picture
Explanation:
<span>It is known
that acids compounds contains hydrogen and produces hydrogen ion in water. A binary
acid however is an acid that have two elements, one of the element has a
hydrogen attached to it. Examples of binary acids are hydrogen fluoride (HF),
hydrogen bromide (HBr) and hydrogen sulfide (H2S). In naming a binary acid, it
has two rules; one, as pure compounds and two, as acid solutions. For pure
compounds, start with the name ‘hydrogen’ and end the anion name with ‘-ide’. For
acidic compounds, start with ‘hydro-‘, end the anion with ‘-ic’ and add ‘acid’.</span>
The answer is 3.
That is if an atom has 5 electrons in its outer shell, then it has 3 unpaired electrons.
As the outer shell have 1 s orbital too and that is fully filled and not available for bonding so it must have 3 unpaired electrons.
So the number of unpaired electrons in an atom that is having 5 electrons in its outer shell is 3.
