Answer:
138 mg
Explanation:
A company is testing drinking water and wants to ensure that Ca content is below 155 ppm (= 155 mg/kg), that is, <em>155 milligrams of calcium per kilogram of drinking water</em>. We need to find the maximum amount of calcium in 890 g of drinking water.
Step 1: Convert the mass of drinking water to kilograms.
We will use the relation 1 kg = 1000 g.

Step 2: Calculate the maximum amount of calcium in 0.890 kg of drinking water

Answer:
All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.
Explanation:
Hope this <em><u>Helped!</u></em> :D
Answer:
Metallic Bonding
Explanation:
Metallic Bonding
In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.
212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.
Answer:
Option A.
Explanation:
Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.
Then, 

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.