Physics
08.04.2023 08:00
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Scientists use either a melting or freezing point apparatus to determine the melting
Scientists use either a melting or freezing point apparatus to determine the melting points of the following unknown compounds as part of their investigation. Based on the melting points, the scientists determine the bond type and record the results in the table as shown.
Which statement correctly identifies the ionic compounds and why?
A.
Compounds 1 and 3 are ionic. Because ionic compounds are composed of oppositely charged ions that are attracted to one another, they require more energy to melt than covalent bonds.
B.
Compounds 2 and 4 are ionic. Because ionic compounds are composed of oppositely charged ions that are attracted to one another, they require more energy to melt than covalent bonds.
C.
Compounds 2 and 4 are ionic. Because covalent compounds are composed of oppositely charged ions that are attracted to one another, they require more energy to melt than ionic bonds.
D.
Compounds 1 and 3 are ionic. Because covalent compounds are composed of oppositely charged ions that are attracted to one another, they require more energy to melt than ionic bonds.
Show Answers
A.
Compounds 1 and 3 are ionic. Because ionic compounds are composed of oppositely charged ions that are attracted to one another, they require more energy to melt than covalent bonds.
B.
Compounds 2 and 4 are ionic. Because ionic compounds are composed of oppositely charged ions that are attracted to one another, they require more energy to melt than covalent bonds.
C.
Compounds 2 and 4 are ionic. Because covalent compounds are composed of oppositely charged ions that are attracted to one another, they require more energy to melt than ionic bonds.
D.
Compounds 1 and 3 are ionic. Because covalent compounds are composed of oppositely charged ions that are attracted to one another, they require more energy to melt than ionic bonds.
ayoismeisjjjjuan
4,5(55 marks)
0.4 A
Explanation:
From the question,
Electric power = Voltage×current
P = VI Equation 1
Make I the subject of the equation
I = P/V Equation 2
Given: P = 96 J/s, V = 230 V.
Substitute into equation 2
I = 96/230
I = 0.4 A.
Hence the current is 0.4 A
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