Answer:
. A closed system allows only energy transfer but no transfer of mass. Example: a cup of coffee with a lid on it, or a simple water bottle. ... In reality, a perfectly isolated system does not exist, for instance hot water in a thermos flask cannot remain hot forever.
Answer:
you didn't ask a question so here is your explanation.
Explanation:
Q = mc∆T. Q = heat energy (Joules, J) m = mass of a substance (kg) c = specific heat (units J/kg∙K) ∆ is a symbol meaning "the change in"
Hello!
Your answer would be polar covalent.
Covalent bonds are where two atoms come together, and share electrons between each other, and are therefore, bonded.
In some cases of molecules that are bonded with a covalent bond, one of the atoms is more, you could call it selfish, and takes more of the electrons. A prime example of this is H20, or water. One of the atoms takes the electrons for longer, and therefore has a more negative charge because electrons are counted as negative charges.
This bond where an atom "hogs" electrons, is called a polar covalent bond, respective to the changing charges for the atoms.
So your answer is d.
Hope this helped!
B. The unknown solution had the lower concentration.
Explanation:
Osmosis is a phenomenon in which the molecules of the solvent has a tendency to move through a membrane which is semipermeable from lower concentrated side to the higher concentration side, so that the concentrations on both sides of the membrane must be equal.
So the unknown solution may have lesser concentration than the isotonic solution so that molecules of that solution move from less concentrated side to the more concentrated side, so its level drops.
<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
