Mass = moles x Mr
So in this case the moles =1
And the Mr equals =11
So you would do - 11x1= 11grams
Answer:
During the recharge process, the water infiltrates at the land's surface and flows through the unsaturated zone. It then crosses the water table and enters the groundwater system. Chloride (NaCl) is important and necessary because the sodium ions are held loosely and exchanged easily with calcium and magnesium ions.
Explanation:
Answer:
(i) 5-day BOD of the waste is 120 mg/l.
(ii) The ultimate carbonaceous BOD (Lo) is 20 mg/l.
Explanation:
The dilution factor D is 0.05.
The initial DO is 8.0 mg/L and the DO after 5 days is 2.0 mg/L.
The BOD of the waste for an unseeded mixture is
The ultimate carbonaceous BOD (Lo) can be calculated as
The given unbalanced equation is as follow,
Sn + H₃PO₄ → H₂ + Sn₃(PO₄)₄
In above equation first let get start with Sn, on left side its one and on right side its 3. So, multiply Sn on left side with 3 to balance Sn, Hence,
3 Sn + H₃PO₄ → H₂ + Sn₃(PO₄)₄
Now, balance Phosphorous, on left side it counts one and on right side it counts 4, so multiply H₃PO₄ on left side with 4 to balance P, Hence,
3 Sn + 4 H₃PO₄ → H₂ + Sn₃(PO₄)₄
Oxygen on both left and right are 16 in number, Hence it is balanced,
Hydrogen on left hand side are 12 in number, while that on right hand side only 2, So, multiply H₂ on right side by 6 to balance hydrogen atoms.
<u>3</u> Sn + <u>4</u> H₃PO₄ → <u>6</u><u> </u>H₂ + Sn₃(PO₄)₄
Above equation is balanced, and coefficients are highlighted bold and underlined.
Answer:
C po Hope it helps salamat
