–4(6x + 3) = -12<span>(x + 10)
</span><span> -24x - 12 = -12x - 120
12x = 108
x = 9
answer
</span><span>B. x = 9</span><span>
</span>
The reaction is missing and it's ;
N2H4 + H2 ---> 2NH3
It asks for the total pressure too.
Answer:
A) Rate of change for NH3 = 152 torr/h
B) The total pressure in the vessel will remain the same.
Step-by-step explanation:
N2H4 + H2 ---> 2NH3
1 mole of N2H4 yields 2 moles of NH3.
From the question, the rate given for N2H4 is 76 torr/h.
Thus, The rate of change for NH3 will be = 2 x 76 = 152 torr/h
Now, on the reaction side, 1 mole of N2H4 reacts with 1 mole of H2. So we have 2 moles on the left hand side.
While on the product side, 2 moles of NH3 are produced.
So the total pressure will remain the same because for every 2 moles on the reaction side, 2 moles are gotten on the product side.
Answer:
x = 10
Step-by-step explanation:
First, to solve for x, you need to isolate the variable, meaning that it needs to be the only term on one side of the equation. To do this, we'll just cancel out the -3 by adding +3 on both sides of our equation.
x - 3 = 7
+3 +3
x = 10
Leaving us with our answer, x = 10
Hope this helped :)
Answer:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Step-by-step explanation:
So we have the expression:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
Recall the order of operations or PEMDAS:
P: Operations within parentheses must be done first. On a side note, do parentheses before brackets.
E: Within the parentheses, if exponents are present, do them before all other operations.
M/D: Multiplication and division next, whichever comes first.
A/S: Addition and subtraction next, whichever comes first.
(Note: This is how the order of operations is traditionally taught and how it was to me. If this is different for you, I do apologize. However, the answer should be the same.)
Thus, we should do the operations inside the parentheses first. Therefore:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
The parentheses is:

Square the 6 and the 4:

Do the operations from left to right. 36 divided by 3 is 12. 12 times 4 is 48:

Therefore, the original equation is now:
![12-[20-2(6^2\div3\times2^2)]\\=12- [20-2(48)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%5C%5C%3D12-%20%5B20-2%2848%29%5D)
Multiply with the brackets:
![=12-[20-96]](https://tex.z-dn.net/?f=%3D12-%5B20-96%5D)
Subtract with the brackets:
![=12-[-76]](https://tex.z-dn.net/?f=%3D12-%5B-76%5D)
Two negatives make a positive. Add:

Therefore:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
True using the formula allows you to find the x intercepts of the equation