Answer:
The correct answer is 8.79 × 10⁻² M.
Explanation:
Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,
No. of moles of NaI = Weight of NaI/ Molecular mass
= 4.11 / 149.89
= 0.027420
The vol. of the solution given is 312 ml or 0.312 L
The molarity can be determined by using the formula,
Molarity = No. of moles/ Volume of the solution in L
= 0.027420/0.312
= 0.0879 M or 8.79 × 10⁻² M
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Answer:

Explanation:
We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.
Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.
We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.
- Na: 22.9897693 g/mol
- C: 12.011 g/mol
- O: 15.999 g/mol
Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.
- Na₂ = 22.9897693 * 2= 45.9795386 g/mol
- O₃ = 15.999 * 3= 47.997 g/mol
- Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol
We will convert using dimensional analysis. Set up a ratio using the molar mass.

We are converting 57.3 grams to moles, so we multiply by this value.

Flip the ratio so the units of grams of sodium carbonate cancel.




The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.
Complete Question:
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
Answer:
2.23x10⁶ g
Explanation:
The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:
0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³
0.7976 g/m³
The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:
A = πR², where R is the radius = 1.01x10² m (half of the diameter)
A = π*(1.01x10²)²
A = 32047 m²
The volume is then:
V = 32047 * 87.32
V = 2.7983x10⁶ m³
The mass of the F⁻ is the concentration multiplied by the volume:
m = 0.7976 * 2.7983x10⁶
m = 2.23x10⁶ g