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>
Answer:
Individual solute particles are broken apart from the solid by the;
c. Solvent
Explanation:
A solution is the homogeneous mixture that is made up of two or more substances formed by dissolving a substance which can be a solid, liquid or gas in another substance known as the solvent which normally the larger part of the fraction of the solution than the solute and can also be a solid, liquid or a gas
In a solution the solvent particles serves to brake of and disperser parts of a solid solute to form a more or less homogeneous mixture
Therefore, the solute particles are broken by the <u>solvent</u> particles in a solution
According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure, T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have

Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa
∴ on rearranging above eq., we get V2 =

∴ V2 = 4 m3
Titration is the method used.
Weathering and chemical substance weathering