- 20 points
Answer:
gathering data
Explanation:
When you do an experiment you want to gather information to make a conclusion on your hypothesis.......-
True
False
Correct Answer: TRUE
The other answer is incorrect, I just did the assignment and got it right.
Using the formula q = mcΔT, and substituting the values for mass, specific heat capacity of iron, and temperature change, it is calculated that it takes approximately 3.058 KJ to warm 125 g of iron from 23.5 °C to 78.0 °C.
To calculate the amount of heatneeded to warm 125 g of iron from 23.5 °C to 78.0 °C, we use the formula q = mcΔT, where 'm' is the mass in kilograms, 'c' is the specific heat capacity, and 'ΔT' is the temperature change. In this case, the mass 'm' is 0.125 kg (since 1 g = 10^-3 kg), the specific heat capacity 'c' of iron is 0.449 J/g°C (or 449 J/kg°C), and 'ΔT' is 78.0 °C - 23.5 °C = 54.5 °C.
Substituting these values into the formula, we get q = (0.125 kg) * (449 J/kg°C) * (54.5 °C), which gives a result of approximately 3.058 KJ.
Therefore, it would take approximately 3,058 KJ to warm 125 g of iron from 23.5 °C to 78.0 °C.
#SPJ6
To warm 125 g of iron from 23.5 °C to 78.0 °C, it requires approximately 3.93 kilojoules of energy.
To calculate the number of kilojoules required to warm 125 g of iron from 23.5 °C to 78.0 °C, we can use the formula:
q = m * c * ΔT
Where:
Using the given values:
Substituting the values into the formula:
q = 125 g * 0.450 J/g°C * (78.0 °C - 23.5 °C)
Simplifying the equation:
q = 125 * 0.450 * (78.0 - 23.5)
q ≈ 3933.75 J ≈ 3.93 kJ
Therefore, it requires approximately 3.93 kilojoules of energy to warm 125 grams of iron from 23.5 °C to 78.0 °C.
#SPJ12
1 mol = 6.022 x 10²³ atoms
In order to find how many atoms, dimly multiply the amount of moles you have by 6.022 x 10²³ or Avogadro's number.
So you have 1.75 mol CHC1₃ x (6.022x10²³) = 1.05385 x 10²⁴ atoms of CHCl₃
But now you have to round because of the rules of significant figures so you get 1.05 x 10²⁴ atoms of CHCl₃
The weighted average mass of the atoms in a naturally occurring sample of an element is the average atomic mass, also known as atomic weight.
What is atomic mass?
The average mass of an element's atoms expressed in atomic mass units is known as its atomic mass (amu, also known as daltons, D).
The mass of each isotope is multiplied by its abundance to produce the atomic mass, which is a weighted average of all the isotopes of that element.
Any element has a variety of isotopes, thus to take into consideration the fact that each isotope has a different mass, a weighted average mass is used.
A weighted average takes into account the potential that an element's average mass, as calculated from a random sample, will most likely coincide with the average mass of its more common isotopes.
The weighted average masswill therefore be closest to that of the more prevalent isotopes.
To learn more about Atomic mass, refer to the below link:
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