<span>The element bromine has two isotopes: Br-79 and Br-81, with a 50%-50% isotopic abundance. Statistically, 25% of bromine molecules will be Br79-Br79, 25% will be Br81-Br81 and 50% will be Br79-Br81. This is equivalent to a ratio of 1:1:2 or 1:2:1. The peaks in a mass spectrum just like chromatography reflect this relative abundance of different isotopic combinations.</span>
This question involves the concepts of density, volume, and mass.
The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".
<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>
It is given that:
Mass of one mole = 24 grams
Mass of 6 x 10²³ atoms = 24 grams
Mass of 1 atom = 
= 4 x 10⁻²³ grams
<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>
where,
= density = 1.7 grams/cm³- m = mass of single atom = 4 x 10⁻²³ grams
- V = volume of single atom = ?
Therefore,
V = 2.35 x 10⁻²³ cm³
<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>
The atom is in a spherical shape. Hence, its Volume can be given as follows:
![V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}](https://tex.z-dn.net/?f=V%20%3D%5Cfrac%7B%5Cpi%20d%5E3%7D%7B6%7D%5C%5C%5C%5Cd%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6V%7D%7B%5Cpi%7D%7D%5C%5C%5C%5Cd%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6%282.35%5C%20x%5C%2010%5E%7B-23%7D%5C%20cm%5E3%29%7D%7B%5Cpi%7D%7D)
d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m
Learn more about density here:
brainly.com/question/952755
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
Answer:
The rise from A to B is 0.887
Solution:
As per the question:
The following reading of an inverted staff is given as:
A = 2.915
B = -2.028
Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.
Thus
The rise from A to B is given as:
A - B = 2.915 - 2.028 = 0.887
