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Marta_Voda [28]
3 years ago
12

If she traveled 10 m in 5 seconds, what was her average speed?

Chemistry
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0
Average speed = distance traveled divided by time taken. so 10 divided by 5, which is 2. her average speed was 2m per second. hope this helps!
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Which of the following best describes an orbital?
Debora [2.8K]

Answer:

a) a space in an atom where an electron is most likely to be found

Explanation:

Atomic orbital is the mathematical function which describes wave-like behavior of the electrons present in the atom.

It is used to calculate probability of finding the electron of the atom in any  region around nucleus of the atom. Atomic orbital is the physical region or a three dimensional space where the probability of finding the electron is more than 90% which also means that the space where an electron is most likely to be found. They are boundless space and have definite energy.

Option A best describes atomic orbital.

6 0
3 years ago
10-The vapor phases of liquids such as acetone and alcohol are more flammable than their liquid phases. For flammable liquids, w
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The relationship between the evaporation rate and the likelihood that the liquid will be flammable is its flash point number (or the lowest temperature for which the liquid ignites) and its activation energy (how much energy is required to ignite it). With volatile liquids or gases these numbers seem to be lower.
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What did the name carbon come from
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3 years ago
Read 2 more answers
Convert 325 milliliters to liters.
steposvetlana [31]

Answer:

The Awnser Is A. 0.325 Liters

Explanation:

6 0
3 years ago
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A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i
Alexeev081 [22]

Answer:

MM_{acid}=140.1g/mol

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}

Thus, solving for the moles of the acid, we obtain:

n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol

Then, by using the mass of the acid, we compute its molar mass:

MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol

Regards.

7 0
3 years ago
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