Answer/Explanation:
Heat from the sun hits the ground and is absorbed. The higher you go, it gets more colder due to the fact that the air doesn't hold onto the radiation as it goes straight through the ground, which is why the top of mountains are very cold, and people are able to die in minutes.
Answer:
Question 2: Na3PO4, KOH; Question 3: Na3PO4, KOH
Explanation:
Question 2
The reactants in a chemical equation are the species on the left side of the reaction arrow.
Thus the reactants are Na3PO4, KOH (sodium phosphate and potassium hydroxide).
Question 3.
The products in a chemical equation are the species on the right side of the reaction arrow.
Thus the products are NaOH, K3PO4 (sodium hydroxide and potassium phosphate).
Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
The unsaturated zone is the portion of the subsurface above the groundwater table. The soil and rock in this zone contains air as well as water in its pores. ... Unlike the aquifers of the saturated zone below, the unsaturated zone is not a source of readily available water for human consumption
Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54