Answer:
....,................................
Explanation:
1= A
2=D
3=C
4=C
Answer: 0.25 mol
Explanation:
Use the formula n=N/NA
n= number of mols
N = number of particles
Nᵃ = Avogadros constant = 6.02 x
So, n=
The 10 to the power of 23 cancels out and you are left with 1.505/6.02, which is approximately 1/4. This is the same as 0.25 mol.
Hope this helped :)
Answer:
- The first picture attached is the diagram that accompanies the question.
- The<u> second picture attached</u> is the diagram with the answer.
Explanation:
In the box on the left there are 8 Cl⁻ ions and 8 Na⁺ ions.
The dissociaton equation for NaCl(aq) is:
- NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)
The dissociation equation for CaCl₂ (aq) is:
- CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)
A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.
Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half) and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).
Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.
Answer:
There was an improvement in accuracy. There was no change in precision.
Explanation:
<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).
The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.
Answer:
C
Explanation:
It afffects changes in pressure and temperature not melting and boiling points