Answer:
0.01144L or 1.144x10^-2L
Explanation:
Data obtained from the question include:
V1 (initial volume) = 20.352 mL
P1 (initial pressure) = 680mmHg
P2 (final pressure) = 1210mmHg
V2 (final volume) =.?
Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:
P1V1 = P2V2
680 x 20.352 = 1210 x V2
Divide both side by 1210
V2 = (680 x 20.352)/1210
V2 = 11.44mL
Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:
1000mL = 1 L
11.44mL = 11.44/1000 = 0.01144L
Therefore the volume of the container is 0.01144L or 1.144x10^-2L
By studying and learning by using your brain to accomplish new goals. :)
(if that makes sense)
Hope this helps :3
Answer: 24.1%, under below assumptions.
Justification:
The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:
- sodium bicarbonate, and
- ammonium bicarbonate
.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.
Given that later, the question is <span>what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.
With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.
In formulas:
100% = %m/m sodium bicarbonate + %m/m ammonium bicarbonate = 100%
=> % m/m sodium bicarbonate = 100% - % m/s ammonium bicarbonate
=> % sodium bicarbonate = 100% - 75.9% = 24.1%
Answer: 24.1%
</span>
Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm
Answer is: The zooplankton can be used to help control algal growth in some coral communities.
Hypothesis is a suggested explanation for an observable phenomenon.
Steps of the scientific method:
1) ask a question about something that is observed.
2) do background research.
3) construct a hypothesis, an attempt to answer questions with an explanation that can be tested. In this example, does the presence of a certain type of zooplankton inhibited the algae growth
4) test of hypothesis.
5) analyze collected data and draw a conclusion. In this example, the zooplankton can be used to help control algal growth.
