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2 years ago

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Calculate the enthalpy of the reaction 2no(g)+o2(g)→2no2(g) given the following reactions and enthalpies of formation: 12n2(g)+o

2(g)→no2(g), δh∘a=33. 2 kj 12n2(g)+12o2(g)→no(g), δh∘b=90. 2 kj.

1 answer:

e-lub [12.9K]2 years ago

5 0

I don’t want you going back in the front door and I’m going to try calling it

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(5372 + 100y^2) + 5(10y^2 – 110) Add the two polynomials and write in the form ay^2 + b, where a and b are constants, What is th

dangina [55]

Answer:

The value of a is 150 and b is 4822.

Explanation:

First, you have to expand to remove brackets :

$(5372 + 100 {y}^{2} ) + 5(10 {y}^{2} - 110)$

$= 5372 + 100 {y}^{2} + 50 {y}^{2} - 550$

Next, you have to simplify :

$100 {y}^{2} + 50 {y}^{2} + 5372 - 550$

$= 150 {y}^{2} + 4822$

Next, you have to compare it :

$a {y}^{2} + b = 150 {y}^{2} + 4822$

$a = 150$

$b = 4822$

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The access code for a card security system consist of four digits the first digit cannot be 4 and the last digit must be odd. Ho

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Different codes for access code which are available are: 4000

_ _ _ _

1 2 3 4

1st blank can be filled with the following possibilities, here the conditions are:

1)The first digit cannot be 4

2)The first digit cannot be 0 also since it makes the access code 3 digits hence

From 1 to 9 except 4 there are 8 possible ways to fill that blank

<u>8</u>

2nd blank can be filled with any number between 0 to 9 in 10 possible ways

<u>8</u> <u>10 </u>

3rd blank can be filled with any number between 0 to 9 in 10 possible ways

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4th blank can be filled with any odd numbers between 0 to 10 - 1,3,5,7,9

<u>8</u> <u>10</u> <u>10</u> <u>5</u> = 8 x 10 x 10 x 5 = 4000

Hence to conclude that 4000 available codes are there for the access security card

To know more about permutations and combinations please follow this link

brainly.com/question/1216161

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