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Nana76 [90]
4 years ago
10

Rubbing alcohol is less polar than water. Both are liquids at room temperature which One boils first? why?

Chemistry
1 answer:
Sergeeva-Olga [200]4 years ago
8 0

Answer:

because both liquid are made from different substances.

Explanation:

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How can a change to another population could affect births and deaths in the parrot population.
forsale [732]

Answer:

read it first

Explanation:

Population change is governed by the balance between birth rates and death rates. If the birth rate stays the same and the death rate decreases, then population numbers will grow. If the birth rate increases and the death rate stays the same, then population will also grow.

hope it helps

6 0
2 years ago
An aqueous solution of 1.29 M ethanol, CH3CH2OH, has a density of 0.988 g/mL. The percent by mass of CH3CH2OH in the solution is
anzhelika [568]

Answer:

6%

Explanation:

Hello, for this case, we assume that the volume of the solution is 1L, thus, the mass is given by using the density as follows:

m_{solution}=1L*\frac{1000 mL}{1L}*\frac{0.988g}{1mL} =988g

Now, the mass of the ethanol:

m_{C_2H_5OH}=(1.29molC_2H_5OH/L*1L)*\frac{46gC_2H_5OH}{1molC_2H_5OH} \\m_{C_2H_5OH}=59.34g

Finally, the by mass percent is:

m/m=\frac{59.34g}{988g}*100\\%

%m=6%

Best regards.

4 0
3 years ago
Ayuda es urgente!! el tronco de madera y las cenizas pesan lo mismo ?​
skad [1K]

Answer:

Spanish

Explanation:

help is urgent the log of wood and the same

3 0
3 years ago
2 KCIO3 - 2 KCl + 302;<br>Type of Reaction?<br><br>​
Ierofanga [76]

Answer:decomposition reaction

Explanation:it is a decomposition reaction

6 0
3 years ago
If a substance has a half life of 58 years and starts with 500 g radioactive, how much remains radioactive after 30 years?
Vilka [71]

Answer:

A = 349 g.

Explanation:

Hello there!

In this case, since the radioactive decay kinetic model is based on the first-order kinetics whose integrated rate law is:

A=Ao*exp(-kt)

We can firstly calculate the rate constant given the half-life as shown below:

k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{58year}=0.012year^{-1}

Therefore, we can next plug in the rate constant, elapsed time and initial mass of the radioactive to obtain:

A=500g*exp(-0.012year^{-1} *30year)\\\\A=349g

Regards!

5 0
3 years ago
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