After finding the oxidation states of atoms, you identify the half reactions (option c).
The half reactions are given by the change of the oxidation states of the atoms.
For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
Cardiovascular and circulatory
kidneys filter thru blood to take out waste
lungs breathe in oxygen, give blood 2 circulatory to carry! takes co2 out
Answer:
1.60x10⁶ billions of g of CO₂
Explanation:
Let's calculate the production of CO₂ by a single human in a day. The molar mass of glucose is 180.156 g/mol and CO₂ is 44.01 g/mol. By the stoichiometry of the reaction:
1 mol of C₆H₁₂O₆ -------------------------- 6 moles of CO₂
Transforming for mass multiplying the number of moles by the molar mass:
180.156 g of C₆H₁₂O₆ ----------------- 264.06 g of CO₂
4.59x10² g ---------------- x
By a simple direct three rule:
180.156x = 121203.54
x = 672.77 g of CO₂ per day per human
So, in a year, 6.50 billion of human produce:
672.77 * 365 * 6.50 billion = 1.60x10⁶ billions of g of CO₂
Answer:
The answer is (D).
Explanation:
A(n) "electromagnet" is an electric circuit that produces a magnetic field.
Answer : The pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
