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2 years ago

15

How do I determine the value for the smallest increment on each cylinder

1 answer:

Inessa05 [86]2 years ago

6 0

You can tell by how many of the smallest increments there are and the first big number. in this case for number 1 each increment would be worth 1, because it reaches ten and has ten increments. While the one next to it would be worth 2.5, because 10 divided by four equals 2.5. Hope this helps!

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A gas system contains 2.00 moles of O2 and CO2 gas, has an initial temperature of 25.0 oC and is under 1.00 atm of pressure. If

Ainat [17]

Answer:

$V_2=52.2L$

Explanation:

Hello there!

In this case, by bearing to to mind the given conditions, it is firstly possible to determine the initial volume of the closed system via the ideal gas equation:

$PV=nRT\\\\V=\frac{2.00mol*0.08206\frac{atm*L}{mol*K}*298.15K}{1.00 atm} \\\\V=48.9L$

Which is V1 in the Charles' law:

$\frac{T_2}{V_2} =\frac{T_1}{V_1}$

And of course, T1 is 298.15 (25+273.15). Therefore, by solving for V2 as the final volume, we obtain:

$V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{48.9L*(45+273.15)K}{(25+273.15)K} \\\\V_2=52.2L$

Best regards!

8 0

3 years ago

1. The common name for the compound CH3-CH2-O-CH3 is

Ket [755]

Answer:

propane

Explanation:

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3 0

2 years ago

Read 2 more answers

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Maurinko [17]

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7 0

2 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:

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

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5 0

3 years ago

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Vlada [557]

Answer:

Explanation:

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4 0

2 years ago