<span>A small proportion of helium in the crust is helium that was trapped in the Earth when the Earth formed and has not yet escaped. Most helium on Earth, however, forms as a result of alpha decay of uranium and thorium. The emitted alpha particles, once they grab a couple of stray electrons, become helium atoms and can accumulate in gas reservoirs along with things such as methane.</span>
Answer:
Following are the responses to the given choices:
Explanation:
- The RBC crenation is implied through NaCl by 2,67 percent(m/v) because that solution becomes hypertonic to RBC because of the water within the RBC that passes externally towards the outskirts. RBC thus shrinks.
- 1.13% (m/v), because the low concentration or osmotic that all this solution shows is hypotonic regarding RBC because of the water which has reached the resulting swelling in RBC.
- Distilled H2 implies hemolytic distillation.
- Glucose is indicated by crenation at 8.69 percent (m/v).
- 5.0% (m/v) glucose and 0.9% (m/v) (Crenation is indicated by NaCl.v)
d, one atom of oxygen and two atoms of hydrogen
. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:
E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ
Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ
2. The wavelength of a photon with this energy would be:
Energy = hc/wavelength
wavelength = hc/energy
wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m
Converting to nanometers gives: 91.16 nm
3. Repeat the calculation in 1, but using n=5.
4. Repeat the calculation in 2 using the energy calculated in 3.
Answer:
4.8 %
Explanation:
We are asked the concentration in % by mass, given the molarity of the solution and its density.
0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:
MW acetic acid = 60.0 g/mol
mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g
mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)
= 1010 g
% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %
