Answer: New concentration of
is 0.23 M.
Explanation:
The given data is as follows.
Moles of
= 0.06 mol
Moles of
= 0.08 mol
Therefore, moles of
added are as follows.
Moles of
= ![0.125 \times \frac{25}{1000}](https://tex.z-dn.net/?f=0.125%20%5Ctimes%20%5Cfrac%7B25%7D%7B1000%7D)
= 0.003125 mol
Now, new moles of
= 0.06 + 0.003125
= 0.063125
Therefore, new concentration of
will be calculated as follows.
Concentration = ![\frac{0.063125}{0.275}](https://tex.z-dn.net/?f=%5Cfrac%7B0.063125%7D%7B0.275%7D)
= 0.23 M
Thus, we can conclude that new concentration of
is 0.23 M.
Answer:
The order of decreasing mass is: I - Fe-Mn- N
Explanation:
According to the periodic table the atomic masses of the elements of the statement are:
IODE (I): 126, 90447
IRON (Fe): 55,845
MANGANESE (Mn): 54, 938844
NITROGEN (N): 14, 0067
The reason as for why there are decimal values for atomic mass values, is because, it is the average of all of the possible atomic masses that can be formed and or present, by taking varying number of neutrons. This varying number of neutrons causes different mass values to be present and by taking the average of the atomic masses, from all possible isotopes, the value yields an average or approximate atomic mass based on all of the atomic masses formed from their unique number of neutrons.
Electrons are not included as they are not found within the nucleus of the atom, for which the atomic mass value can be obtained. They are found outside the Nucleus, within orbitals, of regions of space.
Answer:
Explanation:
From the given information, the rise in a capillary tube h = ![\dfrac{2T cos \theta }{r \rho g}](https://tex.z-dn.net/?f=%5Cdfrac%7B2T%20cos%20%5Ctheta%20%7D%7Br%20%5Crho%20g%7D)
where:
For the height of water:
Surface Tension T = 66.2 mN/m = 66.2 N/m
θ = 10⁰
Cos θ = cos 10 = 0.985
radius r = 0,257 mm = 0.275 × 10⁻³ m
density of water
= 1000 kg/m³
g = 9.8 m/s²
∴
replacing our values:
h = ![\dfrac{2T cos \theta }{r \rho g}](https://tex.z-dn.net/?f=%5Cdfrac%7B2T%20cos%20%5Ctheta%20%7D%7Br%20%5Crho%20g%7D)
h = ![\dfrac{2\times 66.2 \times 10^{-3} \times 0.985}{0.275 \times 10^{-3} \times 1000\times 9.8}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%5Ctimes%2066.2%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%200.985%7D%7B0.275%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%201000%5Ctimes%209.8%7D)
h = 0.0484 meteres
h = 48.3 mm
Since the height h = ![\dfrac{2T cos \theta }{r \rho g}](https://tex.z-dn.net/?f=%5Cdfrac%7B2T%20cos%20%5Ctheta%20%7D%7Br%20%5Crho%20g%7D)
r = 0.275 mm = 0.275 × 10⁻³ m
the density of mecury now
= 13593 kg/m³
the surface tension of the mercury ![T_{Hg} =470 \times 10^{-3} \ N/m](https://tex.z-dn.net/?f=T_%7BHg%7D%20%3D470%20%5Ctimes%2010%5E%7B-3%7D%20%5C%20N%2Fm)
θ = 130⁰
Cos θ = cos 130 = -0.6428
Using the same above equation:
h = ![\dfrac{2\times 470 \times 10^{-3} \times (-0.6428) }{0.275 \times 10^{-3} \times 13593 \times 9.8}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%5Ctimes%20470%20%5Ctimes%20%2010%5E%7B-3%7D%20%5Ctimes%20%28-0.6428%29%20%7D%7B0.275%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%2013593%20%5Ctimes%209.8%7D)
h = -0.016494 m
h = 16.49 mm