The question is incomplete, here is the complete question.
A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.
<u>Answer:</u> The concentration of copper fluoride in the solution is 
<u>Explanation:</u>
To calculate the molarity of solute, we use the equation:
We are given:
Given mass of copper (II) fluoride = 0.0498 g
Molar mass of copper (II) fluoride = 101.54 g/mol
Volume of solution = 100.0 mL
Putting values in above equation, we get:
Hence, the concentration of copper fluoride in the solution is
Answer:
1,869.97 grams of Be(NO3)2
Explanation:
Be(NO3)2 = Be N2 O6
Be=9.012182g/mole
N2=28.0134g/mole
O6=96g/mole
therefore Be(NO3)2 gives you 187.56g in one mole
so 9.97 moles means there is 9.97 times more
9.97mole Be(NO3)2 * 187.56g Be(NO3)2/1mole Be(NO3)2 = 1,869.97g of Be(NO3)2
if you have 1mol of NO. how many molecules of NO are there
Answer:
6.02 x 10²³ molecules
Explanation:
Given parameters:
Number of moles of NO = 1 mole
Unknown:
Number of molecules in NO;
Solution:
A mole of compound contains the Avogadro's number of particles.
1 mole of a substance contains 6.02 x 10²³ molecules
So, 1 mole of NO contained 6.02 x 10²³ molecules
Answer:
It's the button with the x, and a blank box above it.
Explanation:
Check the attachments, you'll see what it looks like. It will probably look different on your calculator, but the icon should be the same or similar.
Answer:
449.5 g
Explanation:
Silver sulfate- Ag2SO4
M(Ag)=107 g/mol => M(Ag2)=214 g/mol
M(S)=32 g/mol
M(O)=16 g/mol => M(O4)=64 g/mol
M(Ag2SO4)=310 g/mol
n=1.45 mol
m(Ag2SO4)=M(Ag2SO4)*n=310 g/mol *1.45 mol= 449.5 g
