Answer: B) Particles can be filtered from a suspension.
Explanation: Colloids are solutions with particle size intermediate between true solutions and suspensions. They exhibit tyndall effect that is scattering of light.
Suspensions have large sized particles which settle when left undisturbed for sometime and thus can be filtered off easily.
The particle size in colloids is less and hence they do not settle under the effect of gravity.
A solution can be homogeneous in which the composition is uniform or heterogeneous in which is the composition is not uniform.
<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
it is most likely a beaker, beacuse 300ml is quite a large volume. Otherwise, it would be a measuring cylinder or pippette
Answer:
I won’t do it in paragraph form cuz it will look very choppy but here you go:
Weathering is when the weather itself changes something, like when a metal bike gets rusty after sitting outside for a long time, or when a plant grows out of concrete.
Erosion is when something gets eroded away at. Like when something has water or wind is flowing against it so much that it changes shape. This is how canyons are made.
Deposition is when a gas turns into a solid, removing energy and skipping the liquid step (frost forming on car window)
Empirical formula is the simplest ratio of whole numbers of components in a compound
in 100 g of compound
C H O
mass 25.5 g 6.40 g 68.1 g
number of moles 25.5 g/12 g/mol 6.40 g/ 1 g/mol 68.1 g/ 16 g/mol
= 2.13 mol = 6.40 mol = 4.26 mol
divide by least number of moles
2.13/2.13 = 1 6.40/2.13 = 3.0 4.26/2.13 = 2.0
all rounded off
C - 1
H - 3
O - 2
empirical formula - CH₃O₂
