Answer:
I needed to use good precision in my measurement for my chemistry lab
Explanation:
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Answer:</h3>
298.15 K
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Explanation:</h3>
W e are supposed to calculate the Value of K at 25°C
Assuming the value of K represent K, the question wants us to convert degree Celsius to Kelvin.
- To convert degrees Celsius to kelvin scale, we use the relationship;
- Kelvin (K) = Degrees Celsius + 273.15 ; 273.5 is a constant
- That is, to convert temperature from °C to Kelvin we add a constant of 273.15 to the °C given.
In this case;
Temperature is 273.15 °c
Thus, to Kelvin scale temperature will be;
= 25°C + 273.15
= 298.15 K
Therefore, the value of K, at 25°C is 298.15 K
Answer:
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Explanation:
According to IUPAC rules, the name of a compound is:
Prefix+root word+suffix
1) Select the longest carbon chain and it gives the root word.
2) The substituents give the prefix.
3) The functional group gives the secondary suffix and the type of carbon chain gives the primary suffix.
The structure of the given compounds are shown below:
Answer:
693K
Explanation:
The enthalpy change in the iron is 3690J
We now apply the formula for enthalpy change which is ΔH=mC∅ where ∅ is the temperature change, m the mass of the substance, and C the specific heat capacity for the substance.
ΔH in this case is 3690J.
Therefore 3690J=21.5g×0.449J/g.K×∅
as we are looking for ∅, we make it the subject of the formula.
∅=3690J/(21.5g×0.44J/g)
∅=390
Temperature=30°C +390
=420°+273
=693K
Answer:
Q = 90,000 J
Explanation:
Given data:
Mass skillet = 2000 g
Specific heat capacity = 0.450 J/g.°C
Energy required to raise temperature = ?
Initial temperature = 25°C
Final temperature = 125°C
Solution:
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 = 125°C - 25°C
ΔT = 100°C
Q = 2000 g × 0.450 J/g.°C × 100°C
Q = 90,000 J
