Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ
Explanation:
Using this formular, q = [mCpΔT] and = [nΔHfusion]
The energy that is needed in the different physical changes is thus:
The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:
q = [mCpΔT]
q = 52.0 x 2.09 x 10
q = 1.09 kJ
While from 0°C to 100°C is calculated as:
q = [mCpΔT]
q = 52.0 x 4.18 x 100
q = 21.74 kJ
And for fusion at 0°C is called Heat of fusion and would be given as:
q = n ΔHfusion
q = 52.0 / 18.02 x 6.02
q = 17.38 kJ
And that required for vaporization at 100°C is called Heat of vaporization and it's given as:
q = n ΔHvaporization
q = 52.0 / 18.02 x 40.7
q = 117.45 kJ
Add up all the energy gives 157.8 kJ
Answer:
The prefix di means 2 and the prefix hexa means 6.
As per IUPAC rules you must use the prefixes to indicate the number of atoms of each element in the formula.
So, the proper name for the compound C2H6 is dicarbon hexahydride.
Explanation:
Mass defect for oxygen-16 = 0. 13261 amu, in the kilograms the mass defect equals to 2.20 × 10⁻²⁸ kg.
<h3>What is mass defect?</h3>
Mass defect is the difference between the mass of of an whole atom and the combined mass of its individual particles present in that atom.
We know that, 1 amu = 1.6 × 10⁻²⁷ kg
Given that, mass defect for oxygen-16 = 0.13261 amu
To calculate this defect in terms of kilograms, we have to convert into kg unit as:
0.13261 amu = 0.13261 amu × 1.6 × 10⁻²⁷ kg/amu
0.13261 amu = 2.20 × 10⁻²⁸ kg
Hence option (2) is correct.
To know more about Mass defect, visit the below link:
brainly.com/question/4334375
Scientists use the physical and chemical properties to help them identify and classify matter. These physical and chemical properties are in a macro-perspective, in which these matter contains compounds, elements and atoms. Hence, matter can be classified in various ways, <span><span>
1. </span>Atomic number either atomic mass each element has</span>
<span><span>2. </span>By substance of that matter either pure substance or mixed substance</span> <span>
3. If they cannot reduce a certain substance into a much smaller quantified atomic structure then they they’ll use (2) to identify and classify it.</span>
A.) CIS/Trans isomers
b.) constitutional isomers
c.) identical
d.) constitutional isomers
e.) identical
