Hello, Thanks for making your post on brainly!
I would say the best answer is C)Mixture
A mixer is made by combining 2 or more elements, but your not making a new element. And there are no chemical reactions happening to make a new element.
Hope this helps. Have a great day.
Answer:
<u>Group 2 is the correct answer.</u>
Explanation:
<u>Chlorine has a charge -1. It means that X is +2 so that the formula is XCl2. </u>
Explanation:
Put the pan into a water-containing dish and mix well. Now use a strainer to transfer the solution into another jar. The salt should disappear in it.
And using a tube with a filtrate, transfer the salts that has sand into another bottle with a filtrate. Therefore the sand is split. Eventually, when all the water vaporizes and the salt stays in the bottle, leave the extra solvent and heat it.
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
Answer:
Covalent
Explanation:
A molecule of C₂H₅OH has C-C, C-H, C-O, and O-H bonds.
A bond between A and B will be ionic if the difference between their electronegativities (ΔEN) is greater than 1.6.

No bond has a large enough ΔEN to be ionic.
C₂H₅OH is a covalent molecule.