The answer for the following problem is mentioned below.
- <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.</em></u>
Explanation:
Given:
mass of calcium phosphate (
) = 125.3 grams
We know;
molar mass of calcium phosphate ( ) = (40×3) + 3 (31 +(4×16))
molar mass of calcium phosphate ( ) = 120 + 3(95)
molar mass of calcium phosphate ( ) = 120 +285 = 405 grams
<em>We also know;</em>
No of molecules at STP conditions(
) = 6.023 × 10^23 molecules
To solve:
no of molecules present in the sample(N)
We know;
N÷
=
N =(405×6.023 × 10^23) ÷ 125.3
N = 19.3 × 10^23 molecules
<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules</em></u>
Answer:
A)The spring scale has a high level of precision and a low level of accuracy.
Explanation:
Hope it works for u guys
The answer is statement #3.
Atomic number shows the number of protons in an atom's nucleus. If the atomic number is 30, then it has 30 protons. The question says it is neutral, so it must have the same amount of protons and electrons.
A is the correct answer based on this.
Answer:
The final dilution is 1:400
Explanation:
Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.
So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400
This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.
