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>
I would say G sorry if it’s not right
Answer:
LOD = 0,0177
LOQ = 0,0345
Explanation:
Detection limit (LOD) is defined as the lowest signal which, with a stated probability, can be distinguished from a suitable blank signal. In the same way, quantification limit (LOQ) is defined as the lowest analyte concentration that can be quantitatively detected with a stated accuracy and precision.
There are many formulas but the most used are:
LOD = X + 3σ
LOQ = X + 10σ
Where X is average and σ is standard desvation
For the blanks readings the average X is 0,0105 and σ is 0,0024
Thus:
<em>LOD = 0,0177</em>
<em>LOQ = 0,0345</em>
I hope it helps!
Answer: Peer-reviewed journal article is the most useful because the information in them had been carefully scrutinized and aproved by people who are experts in that particular field.
Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:
First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?
= 25000 g
Substitute the values into the formula:
= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.
