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2 years ago

Please help me with this question !!!

Chemistry

1 answer:

Art [367]2 years ago

4 0

Density of the vinegar is higher than the density of the oil.

Explanation:

Density of the vinegar is higher than the density of the oil. The consequence of this is that the oil will be the top layer in the pitcher while the vinegar is at the bottom layer in the pitcher.

When mixing oil and vinegar will not produce a mixture because the oil contains non-polar molecules while vinegar is a solution of acetic acid in water and both of them are polar molecules.

Learn more about:

liquids with different densities

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AP CHEM answer and i will give brainliest, image attached

scoundrel [369]

Answer:

what is this i dont know what kind of math this is

6 0

1 year ago

An iron ore was found to contain 92.0% pyrite (fool's gold). Pyrite has the formula FeS2and is 46.5% iron. How many grams of iro

Liula [17]

Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.
50.8x0.08=4.064g
We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start.46.736x0.465=21.73224g
21.73224g+4.064=25.79624g of iron ore 25.8g(3sf)

4 0

2 years ago

Is this correct?<br> Plz answer ASAP!

Angelina_Jolie [31]

YES IT'S @LL CORRECT

3 0

2 years ago

Read 2 more answers

An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.

Tju [1.3M]

Answer:

a) $\Delta S$

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

$\delta S=\frac{\delta Q}{T}$

$\Delta S=\frac{Q}{T}$

The energy balance:

$\delta U=[tex]\delta Q- \delta W$

If the process is isothermical the U doesn't change:

$0=[tex]\delta Q- \delta W$

$\delta Q= \delta W$

$Q= W$

The work:

$W=\int_{V1}^{V2}P*dV$

If it is an ideal gas:

$P=\frac{n*R*T}{V}$

$W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV$

Solving:

$W=n*R*T*ln(V2/V1)$

Replacing:

$\Delta S=\frac{n*R*T*ln(V2/V1)}{T}$

$\Delta S=n*R*ln(V2/V1)}$

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

$\Delta S$

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0

3 years ago

This is actually science but anyways my teacher didn't explain much can someone help?

zepelin [54]

TT homo
Tt hetero
tt homo

5 0

2 years ago