Answer:
<em> = 0.2 mL.</em>
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
=
× 0.010
<em> = 0.2 mL.</em>
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
Answer:
is the isotopic notation of the atom
Explanation:
The isotope notation is:

<em>Where a is the mass number = Number of protons + Number of neutrons</em>
<em>b is atomic number = Number of protons</em>
The atomic number define the nature of the atom, the element with atomic number = 15 is phosphorus, P:

a = 15 protons + 16 neutrons = 31
b = 15
is the isotopic notation of the atom.
Answer:
36.09% is the mass percent composition of calcium in calcium chloride
Explanation:
Mass percent is defined as one hundred times the ratio between the mass of a compound (In this case, Calcium), and the total mass of the sample:
<em>Mass Percent = Mass compound / Total mass of the sample × 100</em>
<em />
Computing the values of the problem:
Mass Percent = 0.690g / 1.912g * 100
Mass percent =
<h3>36.09% is the mass percent composition of calcium in calcium chloride</h3>
Density(D) is defined as Mass(M) divided by Volume(V).
The formula for Density is:
D = M / V.
Another way to remember the formula for Density is to remember "Mass per unit of volume".
I hope this helps!
Explanation:
Starting moles of ethanol acid = 0.020 mol
At the equilibrium 50 % of the ethanol acid molecules reacted
∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %
= 0.010 mol
Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol
Moles of the product
gas formed are calculated as
0.010 mol CH3COOH * 1 mol
/ 2 mol CH3COOH
= 0.005 mol 
Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol 
That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol
Now Calculate the pressure :
0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas
P1/n1 = P2/n2
P2 = P1*n2 / n1
= 0.74 atm * 0.015 mol / 0.020 mol
= 0.555 atm