Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
Answer:
390
Step-by-step explanation:
500*.22=110
500-110=390
390
Answer:
$420
Step-by-step explanation:
To work this out you would need to find the 25% decrease of 560. To do this you would first divide 25 by 100, which gives you 0.25. Then you would minus 0.25 from 1, which gives you 0.75. This is because when finding percentage decreases you would first have to convert it into a decimal. Then you would have to minus it from 1 , to make sure that it will be a 25% decrease not a 75% decrease. Then you would multiply 560 by 0.75, which gives you 420.
1) Divide 25 by 100.

2) Minus 0.25 from 1.

3) Multiply 560 by 0.75.

Answer:
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The pH of the weak acid is 3.21
Butyric acid is known as a weak acid, we need the concentration of [H+] formula of weak acid which is given by this equation :
![[H^{+}]=\sqrt{Ka . Ma}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D%5Csqrt%7BKa%20.%20Ma%7D)
where [H+] is the concentration of ion H+, Ka is the weak acid ionization constant, and Ma is the acid concentration.
Since we know the concentration of H+, the pH can be calculated by using
pH = -log[H+]
From question above, we know that :
Ma = 0.0250M
Ka = 1.5 x 10¯⁵
By using the equation, we can determine the concentration of [H+]
[H+] = √(Ka . Ma)
[H+] = √(1.5 x 10¯⁵ . 0.0250)
[H+] = 6.12 x 10¯⁴ M
Substituting the value of [H+] to get the pH
pH = -log[H+]
pH = -log(6.12 x 10¯⁴)
pH = 3.21
Hence, the pH of the weak acid c3h7cooh is 3.21
Find more on pH at: brainly.com/question/14466719
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