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3 years ago

A standard soft drink can has a volume of 0.355 L. An empty can is sealed while

Chemistry

1 answer:

beks73 [17]3 years ago

8 0

Answer:

0.35 atm

Explanation:

To solve this problem, we use Boyle's Law: $P_1V_1=P_2V_2$ , where P is the pressure and V is the volume.

Here, V_1 = 0.355 L, P_1 = 1.0 atm, and V_2 = 0.125 L. So, just plug these values into the equation:

(1.0) * (0.355) = $P_2$ * (0.125) ⇒ $P_2$ ≈ 0.35 atm

Thus, the pressure is 0.35 atm.

Hope this helps!

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3 years ago

Sakura tested the ph of several solutions. which substances are acidic? check all that apply. cola-flavored soda: 2.6 cranberry

Step2247 [10]

Answer:- Cola-flavored soda:2.6, cranberry juice:2.9 and pickle juice:5.5 are acidic.

Explanations:- As per the pH scale, the solution is acidic if the pH is below 7 and the solution is basic if the pH is above 7. The solution is neutral if the pH is exactly 7.

There are only three solutions with pH value below 7 and so they are acidic. these three solutions are, cola-flavored soda, cranberry juice and pickle juice.

Rest of them are basic as their pH value is greater than 7.

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4 years ago

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When first mixed, concrete can be described as (Blank)

joja [24]

Answer: liquid or fluid
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3 years ago

Which type of perspective makes objects look similar to how they look in real life (select the best answer)

olasank [31]

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3 years ago

An ideal gas at a given initial state expands to a fixed final volume. would the work be greater if the expansion occurs at cons

crimeas [40]

Answer:

Constant pressure

Explanation:

At constant pressure,

$w = -p\Delta V = -p(V_{f} - V_{i})$

At constant temperature,

$w = -RT \ln \left(\dfrac{V_{f}}{V_{i}} \right)$

1 mol of an ideal gas at STP has a volume of 22.71 L.

Let's compare the work done as it expands under each condition from an initial volume of 22.71 L.

Isobaric expansion

$w = -100p(V_{2} - 22.71}); \text{(1 bar$\cdot$L = 100 J)}$

A plot of -w vs V₂ gives a straight line (red) with a constant slope of 100 J/L as in the diagram below (Note that w is work done on the system, so -w is the work done by the system). \

Isothermal expansion

$w= -8.314 \times 273.15 \ln \left(\dfrac{V_{f}}{22.71} \right)\\\\= -2271 \left( \ln V_{f} -\ln22.71 \right)\\= -2271 \left(\ln V_{f} - 3.123 \right)\\= 7092 - 2271\ln V_{f}$

A plot of -w vs V₂ is a logarithmic curve. Its slope starts at 100 J/mol but decreases as the volume increases (the blue curve below).

Thus, the work done during an expansion at constant pressure is greater than if the system is at constant temperature.

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3 years ago