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When a chemical reaction takes place, several molecules combine to produce new ...
Oct 3, 2024
Solution by Steps
step 1
To balance the chemical reaction, we need to equate the number of atoms for each element on both sides of the equation. The equation is: xPb(N3)2+yCr(MnO4)2→pCr2O3+qMnO2+rPb3O4+sNO x \text{Pb}(N_3)_2 + y \text{Cr}(MnO_4)_2 \rightarrow p \text{Cr}_2O_3 + q \text{MnO}_2 + r \text{Pb}_3O_4 + s \text{NO}
step 2
Count the number of atoms for each element on both sides. For the left side: Pb=x,Cr=y,Mn=2y,N=6x,O=8y \text{Pb} = x, \text{Cr} = y, \text{Mn} = 2y, \text{N} = 6x, \text{O} = 8y . For the right side: Pb=3r,Cr=2p,Mn=2q,N=s,O=3p+2q+s \text{Pb} = 3r, \text{Cr} = 2p, \text{Mn} = 2q, \text{N} = s, \text{O} = 3p + 2q + s
step 3
Set up the equations based on the atom counts: 1. x=3r x = 3r (for Pb) 2. y=2p y = 2p (for Cr) 3. 2y=2q 2y = 2q (for Mn) 4. 6x=s 6x = s (for N) 5. 8y=3p+2q+s 8y = 3p + 2q + s (for O)
step 4
The number of independent equations is 5, and the number of variables x,y,p,q,r,s x, y, p, q, r, s is 6. Therefore, we can select 1 variable arbitrarily
Answer
The balanced chemical reaction has 5 independent equations and 6 variables, allowing for 1 variable to be selected arbitrarily.
Key Concept
Balancing chemical equations involves ensuring the number of atoms for each element is equal on both sides of the reaction.
Explanation
In this case, we derived 5 equations from the atom counts, and since there are 6 variables, we can freely choose one variable, which is essential for solving the system.
deal with it with matrix
Generated Graph
Solution by Steps
step 1
We start with the equation for junction A: 85 cars/h enter junction A, of which X1X_1 cars/h head to junction B and the remaining X2X_2 cars/h move towards junction C. Thus, we can express this as: X2=85−X1X_2 = 85 - X_1
step 2
At junction C, an additional 60 cars/h enter. The total cars/h crossing junction C is X2+60X_2 + 60. Out of these, 70 cars/h head to junction D, and the remaining X3X_3 cars/h travel to junction B. Therefore, we can write: X3=(X2+60)−70X_3 = (X_2 + 60) - 70 or X3=X2−10X_3 = X_2 - 10
step 3
We substitute X2X_2 from step 1 into the equation from step 2: X3=(85−X1)−10X_3 = (85 - X_1) - 10, which simplifies to X3=75−X1X_3 = 75 - X_1
step 4
From the information given, we know that of the 70 cars/h that travel towards junction D, 45 cars/h travel to junction B, and the remaining 25 cars/h leave the network. This means that the total cars/h traveling to junction B is X1+X3+45X_1 + X_3 + 45. Setting this equal to the total cars/h entering junction B gives us: X1+(75−X1)+45=85X_1 + (75 - X_1) + 45 = 85
step 5
Simplifying the equation from step 4: X1+75−X1+45=85X_1 + 75 - X_1 + 45 = 85 leads to 120=85120 = 85, which is a contradiction. Thus, we need to check our assumptions and constraints. The values of X1X_1, X2X_2, and X3X_3 must be non-negative and satisfy the flow conditions
Answer
The values of X1X_1, X2X_2, and X3X_3 must be determined under the constraints of the traffic flow, leading to a contradiction in the current setup. Further analysis is needed to find valid values.
Key Concept
Traffic flow analysis involves setting up equations based on the flow of cars through junctions and ensuring conservation of flow.
Explanation
The contradiction indicates that the initial assumptions about the flow may need to be revisited, or additional constraints may need to be considered to find valid values for X1X_1, X2X_2, and X3X_3.
deal with it with matrix
Solution by Steps
step 1
To find all possible combinations of pills that provide exactly the required number of vitamins, we set up the following system of equations based on the table:
1x+1y+0zamp;=5(Vitamin A)2x+1y+1zamp;=13(Vitamin B)4x+3y+1zamp;=23(Vitamin C) \begin{align*} 1x + 1y + 0z & = 5 \quad \text{(Vitamin A)} \\ 2x + 1y + 1z & = 13 \quad \text{(Vitamin B)} \\ 4x + 3y + 1z & = 23 \quad \text{(Vitamin C)} \end{align*}
step 2
We can express this system in matrix form as follows:
[1amp;1amp;02amp;1amp;14amp;3amp;1][xyz]=[51323] \begin{bmatrix} 1 & 1 & 0 \\ 2 & 1 & 1 \\ 4 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 13 \\ 23 \end{bmatrix}
step 3
Solving this system using methods such as substitution or elimination will yield the values of xx, yy, and zz. The solution is found to be:
x=8,y=−3,z=0 x = 8, \quad y = -3, \quad z = 0
step 4
For part (b), we need to minimize the cost function 30x+20y+50z30x + 20y + 50z subject to the constraints from part (a). The constraints are:
x+y+zamp;≥52x+y+zamp;≥134x+3y+zamp;≥23 \begin{align*} x + y + z & \geq 5 \\ 2x + y + z & \geq 13 \\ 4x + 3y + z & \geq 23 \end{align*}
step 5
After evaluating the cost function under the constraints, it is determined that no global minima is found
Answer
For part (a), the solution is x=8x = 8, y=−3y = -3, z=0z = 0. For part (b), no global minima found.
Key Concept
Solving systems of equations and optimization under constraints.
Explanation
The solution involves finding integer values for the number of pills while minimizing costs, subject to vitamin requirements. The negative value for yy indicates that the combination is not feasible under the given constraints.
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