Answer:
<h2>there are 5600000000 micrograms in 5.6 kg</h2>
<h2>........ here your answer </h2>
A bright feild is most commonly used
Answer:
Mass% Cr = 85.5%
Explanation:
<u>Given:</u>
Mass of CrBr3 sample = 0.8409 g
Mass of the AgBr precipitate = 1.0638 g
<u>To determine:</u>
The mass percent of Cr in the sample
<u>Calculation:</u>
The reaction of CrBr3 with silver nitrate results in the precipitation of the bromide ion as silver chloride (AgBr) and Cr as soluble Cr(NO3)2
CrBr3(aq) + 3AgNO3(aq)→ 3AgBr(s) + Cr(NO3)3(aq)
Molecular weight of AgBr =187.77 g/mol
Moles of AgBr precipitated is:
![Moles(AgBr)=\frac{Mass(AgBr)}{Mol.wt(AgBr)}=\frac{0.8409g}{187.77g/mol}=0.004478moles](https://tex.z-dn.net/?f=Moles%28AgBr%29%3D%5Cfrac%7BMass%28AgBr%29%7D%7BMol.wt%28AgBr%29%7D%3D%5Cfrac%7B0.8409g%7D%7B187.77g%2Fmol%7D%3D0.004478moles)
Since 1 mole of AgBr contains 1 mole of Cl, therefore:
# moles of Cl = 0.004478 moles
At wt of Cl = 35.45 g/mol
![Mass(Chloride)=moles*at.wt = .004478moles*34.45g/mol=0.1543](https://tex.z-dn.net/?f=Mass%28Chloride%29%3Dmoles%2Aat.wt%20%3D%20.004478moles%2A34.45g%2Fmol%3D0.1543)
![Mass%(chloride)=\frac{mass(chloride)}{mass(sample)}*100=\frac{0.1543}{1.0638}*100 = 14.50%](https://tex.z-dn.net/?f=Mass%25%28chloride%29%3D%5Cfrac%7Bmass%28chloride%29%7D%7Bmass%28sample%29%7D%2A100%3D%5Cfrac%7B0.1543%7D%7B1.0638%7D%2A100%20%3D%2014.50%25)
Mass%(Cr) = 100 - 14.50=85.5%
1x10^7 equals to 100,00,00 coppers
1x(10^7)millilters
Answer:
0.00757 grams
Explanation:
Find the molar mass of the compound: which is 60.05.
The molar mass is basically just the sum of all the atomic masses of each of the elements.
Then multiply the molar mass by the number of moles in the compound, which is 1.26 x 10^-4 moles.
Your answer should be 0.00757 grams.