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Answer:
W = -262 J.
Explanation:
Hello there!
In this case, according to the given information, we can recall the definition of work in terms of constant pressure and variable volume as follows:

So we plug in the given pressure and volumes to obtain:

Now, we convert this number to J (Pa*m³) by using the shown below conversion factor:

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Most rocks that we encounter in our normal everyday lives are sedimentary rocks. Sedimentary rocks are rocks that have been worn down gradually over long periods of time. Because it takes very long periods of time (couple decades) for these rocks to change, it often seems as if they don't change at all, when in reality the change is too small for us to realize it!
Answer:

Explanation:
We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.
- ΔT= final temperature - initial temperature
The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.
- ΔT= 88.9 °C - 58.8 °C = 30.1 °C
Now we know three variables:
- Q= 4500.0 J
- c= 0.4494 J/g°C
- ΔT = 30.1 °C
Substitute these values into the formula.

Multiply on the right side of the equation. The units of degrees Celsius cancel.

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

The units of Joules cancel.


The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

The mass of the sample of metal is approximately <u>333 grams.</u>
Explanation:
Rutherford conducted an experiment in which he took a thin gold particle film on which he passes alpha- particles. He noticed that:
- Most of the alpha particles get through the film and can be detected by the detector.
- Around small portion of the alpha particle deflected at small angles.
- A very very few alpha particle (approximately 1 out of 1 million alpha particles) just retraced their path which means come back from the center.
He concluded that:
<u>Most of the space of the atom is empty and in the center of the atom , there is solid mass which is the cause of the alpha particles to come back. He gave the term nucleus to this solid mass.</u>