1. A student tried to solve the following problem by selecting the tile as shown. What, if anything, did the student do wrong?
1 answer:
Answer:
B. The student chose the correct tile, but needs to flip the tile to make the units cancel
Explanation:
Based on the reaction:
2AgNO₃(aq) + Cu(s) → 2Ag(s) + Cu(NO₃)₂ (aq)
<em>2 moles of AgNO₃ react per mole of Cu producing 2 moles of Ag and 1 mole of Cu(NO₃)₂</em>
Thus, if you want to produce 6.75moles of Cu(NO₃)₂ you need:
= 13.50 moles of AgNO₃ are needed
Thus, if you analize the tile shown by the student:
<em>B. The student chose the correct tile, but needs to flip the tile to make the units cancel</em>
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Answer:
2.83 g
Explanation:
At constant temperature and pressure, Using Avogadro's law
Given ,
V₁ = 2.12 L
V₂ = 3.12 L
n₁ = 0.120 moles
n₂ = ?
Using above equation as:
n₂ = 0.17660 moles
Molar mass of methane gas = 16.05 g/mol
So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g
<u>2.83 g are in the piston.</u>
Answer:
- Please, find the graph with the labels and points located on the axes in the picture attached.
Explanation:
This is how you meet all the instructions and some important comments to understand how this kind of graphs word:
<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>
The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.
In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.
<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>
A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.
The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).
The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.
<u>3) Now locate and label the points: </u>
- (90, 0.9) ⇒ 90 mb, 0.9 ml
- (100, 0.8) ⇒ 100 mb, 0.8 ml
- (400, 0.2) ⇒ 400 mb, 0.2 ml
- (600, 0.15) ⇒ 600 mb, 0.15 ml
- (760, 0.1) ⇒ 760 mb, 0.1 ml
The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.
So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.
To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.
With that you can now locate each point on your graph.
