The energy required to raise the temperature of 3 kg of iron from 20° C to 25°C is 6,750 J( Option B)
<u>Explanation:</u>
Given:
Specific Heat capacity of Iron= 0.450 J/ g °C
To Find:
Required Energy to raise the Temperature
Formula:
Amount of energy required is given by the formula,
Q = mC (ΔT)
Solution:
M = mass of the iron in g
So 3 kg = 3000 g
C = specific heat of iron = 0.450 J/ g °C [ from the given table]
ΔT = change in temperature = 25° C - 20°C = 5°C
Plugin the values, we will get,
Q = 3000 g × 0.450 J/ g °C × 5°C
= 6,750 J
So the energy required is 6,750 J.
Density gives mass of object per volume...... Here, density is given 8.90 g/cm3 therefore, per cubic centimeter contains 8.90 g Ni. mole of Ni = mass / atomic mass = 8.90 / 58.6934 = 0.1516 mole number of atoms: mole * 6.022 * 10^23 = 0.1516 * 6.022 * 10^23 = 0.9129 * 10^23 = 0.9 * 10^23 (approx.)
The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months
To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Amount remaining (N) = 5%
Original amount (N₀) = 100%
<h3>Number of half-lives (n) =?</h3>
N₀ × 2ⁿ = N
5 × 2ⁿ = 100
2ⁿ = 100/5
2ⁿ = 20
Take the log of both side
Log 2ⁿ = log 20
nlog 2 = log 20
Divide both side by log 2
n = log 20 / log 2
<h3>n = 4.32</h3>
Thus, 4.32 half-lives gas elapsed.
Finally, we shall determine the half-life of the element. This can be obtained as follow.
Number of half-lives (n) = 4.32
Time (t) = 500 years
<h3>Half-life (t½) =? </h3>
t½ = t / n
t½ = 500 / 4.32
t½ = 115.74 years
Multiply by 12 to express in months
t½ = 115.74 × 12
<h3>t½ ≈ 1389 months </h3>
Therefore, the half-life of the radioactive element in months is approximately 1389 months
Learn more: brainly.com/question/24868345
Answer:
The density of water 
.
Explanation:
Density of water = 1.00 g/mL
1 lb = 453.592 g
Density of the water in 
:
The density of water .
