If the battery was removed, the energy produced by the battery would not be able to continue its path along the circuit.
Answer:
1) 950 mL
2) 625 mmHg
3) 426 mL
Explanation:
1) This is the relationship between pressure and volume. This relationship looks like this:
P1*V1 = P2*V2
This means the first pressure times the initial volume is equal to the second pressure times the second volume. We are solving for the second volume. First, convert the mmHg to atm and the mL to L.
1 L * 1 atm = 1.053 atm * X
X = 0.95 L or 950 mL
2) This is the same concept as the last one. :) We don't have to convert the mmHg to atm since the answer wants it in mmHg.
750 mmHg * 0.25 L = 0.3 L * X
X = 625 mmHg
3) The relationship between volume and temperature is similar to the one between pressure and temperature (like the problem in your last question). Remember to convert degrees C to Kelvin and mL to L.
V1 / T1 = V2 / T2
0.4 L / 303 K = X / 323 K
X = 0.426 L pr 426 mL
These problems become much easier once you learn the relationships between the different variables (temp, pressure, volume, etc.) When you have a problem like this, I like to first determine what relationship I am dealing with and then write out what I have and what I am solving for. This helps with organizing the problem. Then just solve it like a normal algebra problem. Always remember to convert temp to Kelvin, mL to L, and pressure to atm (unless it wants it in a different unit, then just make sure all the units match).
Good luck with you studies! :)
The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.
<span>So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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The amount, in mg, of CO present in the room will be 191,520 mg.
<h3>Stoichiometric problem</h3>
The concentration of the gas in the room is 5.7 x 
mg/cm3.
The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.
We can obtain the volume of the room as:
350 x 300 x 320 = 33,600,000 cm3
The concentration is in mg/cm3, meaning that it is mass/volume.
Thus:
mass = concentration x volume = 5.7 x mg/cm3 x 33,600,000 cm3
= 191,520 mg
The mass of CO in the room is 191,520 mg
More on stoichiometric problems can be found here: brainly.com/question/14465605
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