Which element listed below is considered an alkaline earth metal?
1 answer:
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Complete Question:
1. A block of aluminum with a mass of 140 g is cooled from 98.4°C to 62.2°C with a release of 4817 J of heat. From these data, calculate the specific heat of aluminum.
a. Compute the percentage of error to the nearest tenth in the student's calculation if the actual specific heat value for aluminum is 0.9 J/g°C
Answer:
Percent error = 55.56 %
Explanation:
Given the following data;
Mass = 140 grams
Initial temperature = 62.2°C
Final temperature = 98.4°C
Quantity of heat = 4817 Joules.
To find the specific heat capacity;
Heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2- T1
dt = 98.4 - 62.2
dt = 36.2°C
Making c the subject of formula, we have;
Substituting into the equation, we have;
<em>Specific heat capacity, = 0.95 J/g°C</em>
b. To find the percentage error;
Given the following data;
Actual specific heat capacity = 0.9 J/g°C
Experimental specific heat capacity = 0.95 J/g°C
Percent error can be defined as a measure of the extent to which an experimental value differs from the theoretical value.
Mathematically, it is given by this expression;
Substituting into the formula, we have;
<em>Percent error = 55.56 %</em>
Answer:
Explanation:
Conversion problems could be solved using dimensional analysis for convenient use and understanding. Dimensional analysis is a technique of multiplying the measure we have by a fraction corresponding to some two equal measures in different units.
In order to apply dimensional analysis here, we firstly need to know that a prefix 'n' stands for 'nano', and nano means . In this case, we have nanoliters. Using the prefix, we can find a relationship between nanoliter and liter:
We may now apply dimensional analysis. Since we wish to convert into liters, we'll be multiplying the number we have in nL by a fraction that contains this relationship with liters in the numerator and nanoliters in the denominator, so that nL terms cancel out and we would obtain the final answer in liters: