Answer:
34.7mL
Explanation:
First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.
So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn
All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.
So now we need to go to HCl!
To do that we multiply by the molar coefficients in the chemical equation:

This leaves us with 2(0.0956) = 0.1912 mol HCl
Now we use the relationship M= moles / volume , to calculate our volume
Rearranging we get that V = moles / M
Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl
V= 0.0347 L
To change this to milliliters we multiply by 1000 so:
34.7 mL
No of moles of HBr = 3.78g / 80.9 = 0.0467moles
Molarity = Number of moles/ volume. So we have 0.0467 / 4.5 = 0.0103M.
Kw = 1.0 x 10 -14 = [H3O+] [OH-] and [OH-] = 1.0 x 10^(-14) / [H3O+]
So [OH-] = 1.0 * 10^(-14) /1.03 *10^(-2) = 0.9703 * 10^(-12) M
Answer:
clastic, organic, and chemical
Answer:
Kilo- Hecto- Deka- Unit Deci- Centi- Milli
Explanation:
Answer: The correct option is e.
Explanation: In the given question it is asked to tell the element which has 'Sea of electrons' while bonding.
This means that the element should have extra electrons so that it can bond easily or this also means which element can easily donate its electron to form a bond.
- Electronic configuration of Hydrogen =

Hydrogen cannot loose electrons easily because the electron is present in 1s sub-shell which is the closest to the nucleus and hence will be tightly bonded to it.
- Electronic configuration of Helium =

This element will not loose electrons easily because the 1s sub-shell is fully filled and hence will not be available for bonding.
- Electronic configuration of Sulfur =

This element lacks 2 electrons from attaining stable configurations. Hence, it will gain 2 electrons rather than donating for bonding.
- Electronic configuration of Iodine =

This element lacks 1 electron to attain stable configuration. Hence, it will gain 1 electron rather than donating for bonding.
- Electronic configuration of Lithium =

This element has an extra electron in its valence shell and can be easily lost in order to form bond.
Hence, the correct option is e.