Answer:
b. independent/manipulated variable
Explanation:
Independent/manipulated variable - refers to the variable that is changed by the scientist or an experimenter. Only one variable that is independent is required to ensure a fair test in an excellent experiment. As the independent variable is being changed by an experimenter or scientist, data is being recorded simultaneously as they are collected.
This year course engages students in becoming skilled readers of prose written in a variety of periods, disciplines, and
rhetorical contexts and in becoming skilled writers who compose for a variety of purposes. More immediately, the course
prepares the students to perform satisfactorily on the A.P. Examination in Language and Composition given in the spring.
Both their writing and their reading should make students aware of the interactions among a writer’s purposes, audience
expectations, and subjects as well as the way generic conventions and the resources of language contribute to effectiveness
in writing. Students will learn and practice the expository, analytical, and argumentative writing that forms the basis of
academic and professional writing; they will learn to read complex texts with understanding and to write prose of
sufficient richness and complexity to communicate effectively with mature readers. Readings will be selected primarily,
but not exclusively, from American writers. Students who enroll in the class will take the AP examination.
Answer:
1. Exothermic.
2. -1598 kJ.
Explanation:
Hello!
1. In this case, according to the reaction, we can infer that 799 kJ of energy are evolved (given off, released) it means that the enthalpy of reaction is negative as the reactants have more energy than the products; which means this is an exothermic reaction.
2. Here, as we know that the enthalpy of reaction is -799 kJ/mol, we can compute the q-value as shown below, considering the reacted 2 moles of solid iron:
Which means that 1598 kJ of energy are evolved when 2 moles of solid iron react.
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The density of the metal object=6.0
Given:
Volume of the metal object=1.5ml
Mass of the metal object=9.0g
To find:
Density of the metal object
<u>Step by Step Explanation:
</u>
Solution:
According to the formula, Density of the metal object can be calculated as
Where, m=mass of the metal object
=density of the metal object
v=volume of the metal object
We know the values of v=1.5ml and m=9.0g
Substitute these values in the above equation we get
=9.0/1.5
=6.0
Result:
Thus the density of the metal object is 6.0
