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3 years ago

What r some important things in this paragraph ????

2 answers:

6 0

Keeping your arm straight in front of you, you rotate 90° to your left, and see the left side of the circle lit while the right side is dark. Half the ball is still lit up, but you can see only part of the lit area. As you continue to rotate, you see a different amount of the ball.

stich3 [128]3 years ago

3 0

You can explain how the moon changes in a pattern using models. Suppose you were standing in the dark room with a bare light bulb behind you. You hold a ball in front of you, and you can see all of the lit half of the ball, which looks like a circle. Keeping your arm straight in front of you, you rotate 90° to your left, and see the left side of the circle lit while the right side is dark. In fact, you would see the shape change from a full circle to a crescent shape to a backwards crescent shape and then back to a full circle.

Try this :)

You might be interested in

Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: when you answer the question,

vodomira [7]

Answer:

Explanation:

The result will be affected.

The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).

Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.

During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.

Thus, NaOH will appear at a higher concentration than it actually is.

7 0

4 years ago

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

4 years ago

A gas occupies 56 L at 73°C. What volume will the gas occupy if the temp. cools to 0°C?

vova2212 [387]

Answer:

44.2 L

Explanation:

Use Charles Law:

$\frac{V1}{T1} =\frac{V2}{T2}$

We have all the values except for V₂; this is what we're solving for. Input the values:

$\frac{56 L}{346K} =\frac{V2}{273K}$ - make sure that your temperature is in Kelvin

From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:

$\frac{56*273}{346} = V2$

Therefore, V₂ = 44.2 L

It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.

7 0

3 years ago

What is the concentration of a solution?

Nimfa-mama [501]

The concentration of a substance is the quantity of solute present in a given quantity of solution.

8 0

3 years ago

Classify each amino acid into the following categories. Select the amino acid that fits best in each category. Each amino acid w

mel-nik [20]

Answer:

1. Lysine

2. Aspartic acid

3. Serine

4. Alanine

5. Tryptophan

Explanation:

Amino acids are biomolecules that contain two functional groups and one R side chain. The two functional groups are: carboxyl group and amino group.

The α-amino acids are the amino acids in which the two functional groups and the R side chain are attached to the α-carbon of the amino acid. They are total 22 α-amino acids.

1. A basic amino acid: Lysine is a positively charged, polar basic amino acid with a lysyl side chain.

2. An acidic amino acid: Aspartic acid is a negatively charged, polar acidic amino acid with an acidic carboxymethyl group.

3. A neutral polar amino acid: Serine is a polar and neutral amino acid with a hydroxymethyl group.

4. A non-polar aliphatic amino acid: Alanine is an aliphatic, nonpolar and neutral amino acid with a methyl side chain.

5. An aromatic amino acid: Tryptophan is an aromatic, nonpolar and neutral amino acid with an indole side chain.

3 0

3 years ago