A meteorologist predicts fog in the morning over lakes and river bottoms because of the atmospheric conditions and their interactions with the <span>hydrosphere. The hydrosphere is the region on the Earth's surface where the total water on the planet is found. The hydrosphere can be in the form of liquid, vapor, or ice.</span>
The coefficient in a chemical formula represents the amount of each chemical present. The amount of a substance is measured in moles.
Answer:
1.022ppm is the unknown concentration of the metal
Explanation:
Based on Lambert-Beer law, the increasing in signal of a detector is directly proportional to its concentration.
The unknown concentration (X) produces a signal of 0.255
99mL * X + 1mL * 100ppm / 100mL produces a signal of 0.502
0.99X + 1ppm produce 0.502, thus, X is:
0.255 * (0.99X + 1 / 0.502) =
X = 0.503X + 0.508
0.497X = 0.508
X =
1.022ppm is the unknown concentration of the metal
Answer:

Explanation:
Hello,
In this case, since the chemical reaction is:

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:
![[H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BHCl%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-1.25%7D%3D0.0562M)
Then, since the concentration and the volume define the moles, we can write:
![[HCl]*V_{HCl}=2*n_{Mg(OH)_2}](https://tex.z-dn.net/?f=%5BHCl%5D%2AV_%7BHCl%7D%3D2%2An_%7BMg%28OH%29_2%7D)
Therefore, the neutralized volume turns out:

Best regards.
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g