Answer:
28.75211 kj
Explanation:
Given data:
Mass of iron bar = 841 g
Initial temperature = 84°C
Final temperature = 7°C
Heat released = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
specific heat capacity of iron is 0.444 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 7°C - 84°C
ΔT = -77°C
By putting values,
Q = 841 g × 0.444 j/g.°C × -77°C
Q = 28752.11 j
In Kj:
28752.11 j × 1 kJ / 1000 J
28.75211 kj
Answer: The final temperature of both the weight and the water at thermal equilibrium is 
.
Explanation:
The given data is as follows.
mass = 7.62 g, 
Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.
q = 
= 
Now, heat gained by lead will be calculated as follows.
q = 
= 
According to the given situation,
Heat lost = Heat gained
=
T =
Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is .
The order would be coefficient, law of conservation of mass, products, and reactants respectively.
<h3>Word matching</h3>
The number written in front of a chemical symbol in an equation is called a coefficient.
The total mass of a system being unchanged is known as the law of conservation of mass
The substances made in chemical reactions are called products.
The starting materials in chemical reactions are called reactants.
More on reactions can be found here: brainly.com/question/17434463
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If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.
In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.
So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
Answer:
Non-polar covalent bond
Explanation:
The bond between two nonmetal atoms is always covalent.
