In order to find out the %mass dolomite in the soil,
calculate for the mass of dolomite using the information given from the
titration procedure. You would need to multiply 57.85 ml with 0.3315 M HCl and
you would get the amount of HCl in millimoles. Then multiply the amount of HCl
with 1/2 (given that for every 1 mol of dolomite, 2 mol of HCl would be
needed). Convert the amount of dolomite to mass by multiplying the millimoles
with the molecular weight which is 184.399. Then convert the mass to grams
which is 1.768 grams. Divide the mass of dolomite (1.768 grams) with the weight
of soil sample. The % mass is 7.17.
<em><u>the</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u> </u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>aluminium</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>1</u></em><em><u>4</u></em>
Atomic mass Ca = 40 a.m.u
1 mole Ca ----------- 40 g
2.5 mols Ca -------- ( mass Ca )
Mass Ca = 2.5 x 40 / 1
Mass Ca = 100 / 1
= 100 g of Ca
hope this helps!
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is 
Answer:PLEASE MARK BRAINIEST
The most common method astronomers use to determine the composition of stars, planets, and other objects is spectroscopy. Today, this process uses instruments with a grating that spreads out the light from an object by wavelength. This spread-out light is called a spectrum. Every element — and combination of elements — has a unique fingerprint that astronomers can look for in the spectrum of a given object. Identifying those fingerprints allows researchers to determine what it is made of.
That fingerprint often appears as the absorption of light. Every atom has electrons, and these electrons like to stay in their lowest-energy configuration. But when photons carrying energy hit an electron, they can boost it to higher energy levels. This is absorption, and each element’s electrons absorb light at specific wavelengths (i.e., energies) related to the difference between energy levels in that atom. But the electrons want to return to their original levels, so they don’t hold onto the energy for long. When they emit the energy, they release photons with exactly the same wavelengths of light that were absorbed in the first place. An electron can release this light in any direction, so most of the light is emitted in directions away from our line of sight. Therefore, a dark line appears in the spectrum at that particular wavelength.
Explanation: