Answer: concentration
Explanation:
Concentration refers to the amount of a substance present in a sample. The more molecules of a substance present in a sample, the greater its concentration. The less molecules of a substance in a sample, the lesser the concentration. We are often concerned about analytically determining the concentration of a substance using diverse analytical methods in chemistry.
Answer:
True rod cell are best in the dark or dim light.
Answer:
95,000 centigrams
Explanation:
There is 1000 CG in 0.01 kilograms
so you do 1000*95 which equals 95,000 centigrams.
<h3>
Answer:</h3>
6.25 atoms
<h3>
Explanation:</h3>
<u>We are given</u>;
- The half life of Po-218 is 3 minutes
- Initial sample is 200 atom
- Time of decay is 15 minutes
We are required to calculate the remaining mass after decay;
Half life refers to the time taken for original amount of a radioactive sample to decay to a half.
To calculate the remaining mass we use the formula;
N = N₀ × 0.5^n where n is the number of half lives, N is the new amount and N₀ is the original amount.
n = 15 min ÷ 3 min
= 5
Therefore;
New amount = 200 atom × 0.5^5
= 6.25 atoms
Therefore; the amount of the sample that will remain after 15 minutes is 6.25 atoms.
63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
The balanced reaction equation is;
2 Na + 3 N2--> 2 NaN3
Number of moles of Na = 50.0 g/23 g/mol = 2.17 moles of Na
Number of moles of Nitrogen = 40.5 g/28 g/mol = 1.45 moles of N2
We have to obtain the limiting reactant, this is the reactant that yields the least number of moles of product.
For Na
2 moles of Na yields 2 moles of NaN3
2.17 moles of Na yields 2.17 moles of NaN3 (reaction is 1:1).
For N2
3 moles of N2 yields 2 moles of NaN3
1.45 moles of N2 yields 1.45 * 2/3 = 0.97 moles of NaN3
So, N2 is the limiting reactant. Mass of product formed depends on the limiting reactant.
Mass of NaN3 = 0.97 moles of NaN3 * 65 g/mol = 63.1 g of NaN3
Therefore, 63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
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