Metals :-
Group 1A - Alkali metals ( highly reactive metals)
Non-metals :-
Group 17 - Halogens ( highly reactive non-metals )
Answer:
70.0°C
Explanation:
We are given;
- Amount of heat generated by propane as 104.6 kJ or 104600 Joules
- Mass of water is 500 g
- Initial temperature as 20.0 ° C
We are required to determine the final temperature of water;
Taking the initial temperature is x°C
We know that the specific heat of water is 4.18 J/g°C
Quantity of heat = Mass × specific heat × change in temperature
In this case;
Change in temp =(x-20)° C
Therefore;
104600 J = 500 g × 4.18 J/g°C × (x-20)
104600 J = 2090x -41800
146400 = 2090 x
x = 70.0479
=70.0 °C
Thus, the final temperature of water is 70.0°C
0.003 moles of NaOH was used in the titration.
<h3>What is titration?</h3>
The concentration of an identified analyte can be found using a simple laboratory technique called titration. As a standard solution with a given concentration and volume, a reagent known as the titrant or titrator is created.
By using a solution with a known concentration to measure the concentration of an unknown solution, this process is known as titration. To a known volume of the analyte (the unknown solution), the titrant (the known solution) is typically added from a buret until the reaction is finished. To ascertain the unknown concentration of an identifiable analyte, titration, commonly referred to as titrimetry, is a widely used quantitative laboratory analytical technique (Medwick and Kirschner, 2010). Volume measurements are a crucial component of titration
Concentration in mol/dm3 =
Amount of solution mol
= concentration in mol/dm3 × volume in dm3
Amount of sodium hydroxide
= 0.100 × 0.0250
= 0.00250 mol
To know more about titration, visit:
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Answer:
29.575%
Explanation:
Data provided:
Calories taken in daily diet = 2000
Recommended amount of fat = 65 grams
Average number of calories for fat = 9.1 calories / g
Thus,
Number of calories in the diet with average number of calories for fat
= Recommended amount of fat × Average number of calories for fat
= 65 × 9.1
= 591.5 calories
Therefore,
the percentage of calories in his diet supplied = ( 591.5 / 2000 ) × 100
= 29.575%