Answer:
7.32 g of F₂
Solution:
The equation is as follow,
2 LiI + F₂ → 2 LiF + I₂
According to equation,
51.88 g (2 mole) of LiF is produced from = 37.99 g (1 mole) F₂
So,
10 g of LiF will be produced by = X g of F₂
Solving for X,
X = (10 g × 37.99 g) ÷ 51.88 g
X = 7.32 g of F₂
Answer:
Me and my friends were going to do a science experiment. Jonny’s job was to make the HYPOTHESIS. He said the “ If we mix baking soda and vinegar together, the TEMPERATURE will go down.”
So then Molly mixed the baking soda and vinegar together and checked the TEMPERATURE. We all OBSERVED as the thermometer’s TEMPERATURE went down. “ your THEORY/ HYPOTHESIS was correct!” Exclaimed Molly.
Then the whole science GROUP let out with a cheer! And wrote the information down on their EXPERIMENTAL info chart. They took a microscope and looked at the mixture because they wanted to the the little PARTICLES in the mixture. Lily CONTROLED the microscope she zoomed in and out to see the particles.
Explanation:
i hope this helps:)
Answer: 1.997 M
Explanation:
molarity = moles of solute/liters of solution or 
first we have to find our moles of solute (mol), which you can find by dividing the mass of solute by molar mass of solute
mass of solute: 92 g
molar mass of solute: 46.08 g/mol
let's plug it in:
next, we plug it into our original equation:
Answer:
Demo Mole Quantities
58.5g NaCl(mol/58.5g)(6.02 x 1023/mol) = 6.02 x 1023 Na
+
Cl21 pre-1982 pennies (after 1982 pennies are mostly zinc with copper coating)
63.5g Cu( mol/ 63.5g)(6.02 x 1023/mol) = 6.02 x 1023 Cu
19.0g Al (mol/27.0g)(6.02 x 1023/mol) = 4.24 x 1023 Al
Explanation:
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:
Finally, the resulting molarity in 30.8 mL (0.0308 L):
Regards.
