V ( HCl ) = 16.4 mL / 1000 => 0.0164 L
M( HCl) = ?
V( KOH) = 12.7 mL / 1000 => 0.0127 L
M(KOH) = 0.620 M
Number of moles KOH:
n = M x V
n = 0.620 x 0.0127
n = 0.007874 moles of KOH
number of moles HCl :
<span>HCl + KOH = H2O + KCl
</span>
1 mole HCl ------ 1 mole KOH
<span>? mole HCl--------0.007874 moles KOH
</span>
moles HCl = 0.007874 * 1 / 1
= 0.007874 moles of HCl
M = n / V
M = 0.007874 / <span>0.0164
</span>= 0.480 M
Answer (2)
hope this helps!
Answer:- 448 mL of hydrogen gas are formed.
Solution:- It asks to calculate the volume of hydrogen gas formed in milliliters at STP when 0.020 moles of magnesium reacts with excess HCl acid. The balanced equation is:
There is 1:1 mol ratio between Mg and hydrogen gas. So, the moles of hydrogen gas is also equals to the moles of Mg reacted.
moles of Hydrogen gas formed = 0.020 mol
At STP, volume of 1 mol of the gas is 22.4 L. We need to calculate the volume of 0.02 moles of hydrogen gas.
= 0.448 L
They want answer in mL. So, let's convert L to mL using the conversion formula, 1L = 1000mL
= 448 mL
So, 0.020 moles of magnesium would produce 448 mL of hydrogen gas at STP on reacting with excess of HCl acid.
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Answer:
\left \{ {{y=206} \atop {x=82}}Pb \right.
Explanation:
isotopes are various forms of same elements with different atomic number but different mass number.
Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are
- Alpha particle emission \left \{ {{y=4} \atop {x=2}}He \right.
- Beta particle emission \left \{ {{y=0} \atop {x=-1}}e \right.
- gamma radiation \left \{ {{y=0} \atop {x=0}}γ \right.
in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.
Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below
\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right + \left \{ {{y=0} \atop {x=0}}γ\right.
Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right
M(P)=3.72 g
M(P)=31 g/mol
m(Cl)=21.28 g
M(Cl)=35.5 g/mol
n(P)=m(P)/M(P)
n(P)=3.72/31=0.12 mol
n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol
P : Cl = 0.12 : 0.60 = 1 : 5
PCl₅ - is the empirical formula
