Answer:
B- Sodium loses an electron.
D- Fluorine gains an electron.
Sodium is oxidized.
Explanation:
The reaction equation is given as:
Na + F → NaF
In this reaction, Na is the reducing agent. It loses an electron and then becomes oxidized. By so doing, Na becomes isoelectronic with Neon.
Fluorine gains the electron and then becomes reduced. This makes fluorine also isoelectronic with Neon.
This separation of charges on the two species leads to an electrostatic attraction which forms the ionic bonds.
Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
Add an alkaline compound to raise the pH to 7.2.
Answer:
Scandium
Titanium
Vanadium
Chromium
Manganese
Iron
Cobalt
Nickel
Copper
Zinc
Yttrium
Zirconium
Niobium
Molybdenum
Technetium
Ruthenium
Rhodium
Palladium
Silver
Cadmium
Lanthanum
Hafnium
Tantalum
Tungsten
Rhenium
Osmium
Iridium
Platinum
Gold
Mercury
Actinium
Rutherfordium
Dubnium
Seaborgium
Bohrium
Hassium
Meitnerium
Darmstadtium
Roentgenium
Copernicium
Explanation:
all of those are transition metals lol
2C3H8+ 702--->6CO2+8H20
FROM Equation above 2 moles of C3H8 reacted with 7 moles of oxygen to form 6 moles of c02 plus 8 molesof H2O
the moles of c3H8 reacted is = MASS/ R.F.M
THE R.F.M =48+8=44
Number of moles is hence 0.025/44=5.68x10^-4
since ratio of C3H8 to O2 is 2:7 Therefore moles of O2 reacted is 1.989 x10^-3
mass= r.f.m x number of moles
(1.989x10^-3) x 32 =0.064g
