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user100 [1]
3 years ago
9

Sound Wave A is moving through a medium that is decreasing in temperature. Sound Wave B is moving from a liquid to a gas medium.

Which statement can be made about both waves? They will speed up. They will increase the density of their medium. They will slow down. They will decrease the density of their medium.
Chemistry
2 answers:
Monica [59]3 years ago
4 0

Answer: I think the first guy to answer is correct

Explanation:

olganol [36]3 years ago
3 0

Answer: they will slow down

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Balance the following chemical equation:<br> C2H60+02 → CO2 + H20
pantera1 [17]

Answer:

Explain: In order to balance the chemical equation, you need to make sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. In order make both sides equal, you will need to multiply the number of atoms in each element until both sides are equal.

3 0
2 years ago
In Denver, Colorado the elevation is about 5,280 feet above sea level. Explain what potential effects this may have on the solub
Colt1911 [192]

Answer:

The solubility of the gaseous solute decreases

Explanation:

As we know, pressure decreases with altitude. This means that, at higher altitudes, the pressure is much lower than it is at sea level.

The solubility of a gas increases with increase in pressure and decreases with decrease in pressure.

Hence, in Denver, Colorado where the elevation is about 5,280 feet above sea level, a gaseous solute is less soluble than it is at sea level due to the lower pressure at such high altitude.

4 0
3 years ago
A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
2 years ago
I need help as soon as possible
Pachacha [2.7K]

Answer:

D

Explanation:

3 0
3 years ago
Read 2 more answers
What unknown quantity can be calculated after performing a titration? a. volume c. mass b. concentration d. density
sertanlavr [38]
C & B are switched so I'm not sure if that was a typo or not, but the answer is concentration!
3 0
3 years ago
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