Answer:
The electrode that removes ions from solution
Explanation:
Each electrochemical cell consists of an anode and a cathode. Oxidation occurs at the anode and reduction occurs at the cathode.
At the anode, ions move from the electrode into the solution while at the cathode ions move from the solution to the electrode.
At the cathode, metal ions accept electron(s) and become deposited on the electrode hence this electrode removes ions from solution. This is reduction.
<span>1.40 x 10^5 kilograms of calcium oxide
The reaction looks like
SO2 + CaO => CaSO3
First, determine the mass of sulfur in the coal
5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4
Now lookup the atomic weights of Sulfur, Calcium, and Oxygen.
Sulfur = 32.065
Calcium = 40.078
Oxygen = 15.999
Calculate the molar mass of CaO
CaO = 40.078 + 15.999 = 56.077
Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight.
8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles
Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass
2.49 x 10^3 * 56.077 = 1.40 x 10^5
So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
Answer:
Explanation:
2 HCl(g) + Mg(s) → MgCl₂(s) + H₂(g)
Let's calculate the quantity of mole of produced hydrogen with the Ideal Gases Law
P . V = n . R .T
2.19 atm . 6.82L = n . 0.082 . 308K
(2.19 atm . 6.82L) / (0.082 . 308K) = n
0.591 mol = n
1 mol of H₂ gas came from 2 mol of hydrochloric, so, 0.591 mol came from the double of mole
0.591 .2 = 1.182 mole of acid.
Molar mass of HCl = 36.45 g/m
1.182 mole are (36.45 g/m . 1.182g ) contained in 43.1 g
Density HCl = HCl mass / HCl volume
0,118 g/mL = 43.1 g / HCl volume
43.1 g / 0.118 g/mL = 365.3 mL (HCl volume)
An individual is hospitalized and the initial blood work indicates high levels of
in the blood and a pH of 7. 47. This would indicate the individual probably has compensated respiratory acidosis.
A chronic illness usually leads to compensated respiratory acidosis because the kidneys have time to adjust to the delayed onset. Even if the
is elevated in a compensated respiratory acidosis, the pH is within the usual range.
The kidneys counteract a respiratory acidosis by increasing the amount of
that tubular cells reabsorb from the tubular fluid, the amount of
that collecting duct cells secrete while also producing
, and the amount of
buffer that is formed through ammoniagenesis.
Respiratory acidosis is frequently brought on by hypoventilation as a result of: breathing depression , paralysis of the respiratory muscles, diseases of the chest wall , abnormalities of the lung parenchyma and abdominal squeezing.
Learn more about Respiratory acidosis here;
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