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>
The answer is an igneous rock.
Hope this helps!!
It's 2, glass. Water, nitrogen, and sucrose don;t have a crystalline structure.
Answer:
break it down and just put the numbers
Explanation:
Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.