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Lunna [17]
2 years ago
8

Determine whether the graph is that of a function by using the vertical-line test. If it is, use the graph to find (a) its domai

n and range (b) the intercepts, if any (c) any symmetry with respect to the x-axis, y-axis, or the origin​

Chemistry
1 answer:
Ilia_Sergeevich [38]2 years ago
3 0

Answer:

Following are the solution to this question:

Explanation:

If the draw a line perpendicular with y-axis thru the diagonal line check but it meets only one curved point, therefore the curve indicates a function not otherwise.  They draw a vertical line perpendicular to the y-axis there, it just intersects one more chart point, which is why a graph is a feature:

In point a:

\to Domain= [-\pi ,\pi ]\\\\\to Range = [-1,1]

In point b:

Its y-axis length cut also by understanding the benefits of y-interception and the x-axis length gives the x-intercept.  

\to y-intercept = 1\\\\\to x-intercept = \frac{\pi}{2} ,- \frac{\pi}{2} \\\\

In point C:

Every graph is y-axis symmetric because the left side of the column as well as the middle side of the graph is about the same.

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Calculate the number of grams contained in each of the following number of moles. Report your answer with the appropriate number
Debora [2.8K]

<u>Answers:</u>

a. 131.85 grams

b. 134.69 grams

To calculate the number of grams in the number of moles given for each compound, consider the formula:

<em>Moles = No. of grams / molar mass</em>

Therefore, to find the gram it will become:

<em>No. of grams = moles x molar mass</em>

a.  0.410 mol of NaCl (sodium chloride)

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No. of grams = moles x molar mass

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b. 4.80 mol of C2H4 (ethylene)

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3 years ago
As shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures lis
Setler79 [48]
Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2

when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
                = 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
 it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
         = 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value) 
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.

c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
       =  13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.


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3 years ago
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