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Temka [501]
3 years ago
14

Un aerosol contine 350 ml de gas en su interior . Su presion es de 5 atm. Un estudiante coloca un arosol en una bolsa plastica y

lo oprime para que salga todo el gas la bolsa se infla hasta completar 2.5 L. Que presion hay en el interior de la bolsa
Chemistry
1 answer:
Elenna [48]3 years ago
6 0

Answer:

,b dksb dfsj kdhfsu dfsh khfs fshk

Explanation:

You might be interested in
If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
Mechanism with a small activation energy or one with large activation energy​
tangare [24]

Answer:

Rate depends on the rate constant. The rate constant depends on temperature and activation energy. If you have lower activation energy the rate will be higher. This is why catalysts are added since catalysts provide an alternate pathway that requires lower activation energy and catalysts are added to increase the rate of reaction.

Explanation:

This is only the answer if you were asking:

"Which corresponds to the faster rate: a mechanism with a small activation energy or one with a large activation energy?"

Thats what I understood about your question.

8 0
3 years ago
3 points
VLD [36.1K]

Answer:

<h3>The answer is 8.29 %</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual density = 19.30g/L

error = 20.9 - 19.3 = 1.6

We have

p(\%) =  \frac{1.6}{19.3}  \times 100 \\  = 8.290155440...

We have the final answer as

<h3>8.29 %</h3>

Hope this helps you

5 0
3 years ago
What is the percent composition by mass of nitrogen in NH4OH?
Juliette [100K]
 <span>the atomic mass of nitrogen is 14. There is 1 nitrogen atom in the molecule so the percentage of N is :

14/35 x100% = 40%</span>
8 0
3 years ago
Actual value= -273 Experimental value= -274
Maurinko [17]

Answer:

<h2><u><em>When measuring data, the result often varies from the true value. Error can arise due to many different reasons that are often related to human error, but can also be due to estimations and limitations of devices used in measurement. Regardless, in cases such as these, it can be valuable to calculate the percentage error. The computation of percentage error involves the use of the absolute error, which is simply the difference between the observed and the true value.</em></u></h2>

Explanation:

4 0
3 years ago
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