Answer:
12 %
Explanation:
Produces 15 J of work for input 125 J
15/125 * 100% = 12%
Answer:
a. alkyne
b. alkane
c. alkyne
d. alkene
Explanation:
The general formula for each class of compound is given below
Alkane: 
Alkene: 
Alkyne: 
(assuming single multiple bonds)
Now let us classify according to the above formulas:
a. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
b. It has two hydrogen atoms more than the two times of carbon atoms hence, it's alkane
c. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
d. It has hydrogen atoms two times of carbon atoms hence, it's alkene
Answer:
The First choice is correct
Explanation:
That is the closest example of what is shown
<span>The student is incorrect because helium has 2 valence electrons and it's in group 18 because the first energy level is full. Although helium is placed in Group 18 which generally has 8 valence electrons, it does not have 8 valence electrons as the student suggested. It was grouped together with the noble gases because it exhibits similar properties with them. </span>
Answer:
The solution is basic.
Explanation:
We can determine the nature of the solution via determining which has the large no. of millimoles (acid or base):
- If no. of millimoles of acid > that of base; the solution is acidic.
- If no. of millimoles of acid = that of base; the solution is neutral.
- If no. of millimoles of acid < that of base; the solution is basic.
- We need to calculate the no. of millimoles of acid and base:
no. of millimoles of acid (HNO₃) = MV = (1.3 M)(75.0 mL) = 97.5 mmol.
no. of millimoles of base (NaOH) = MV = (6.5 M)(150.0 mL) = 975.0 mmol.
<em>∴ The no. of millimoles of base (NaOH) is larger by 10 times than the acid (HNO₃).</em>
<em>So, the solution is: basic.</em>
