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Luba_88 [7]
3 years ago
8

1.)­ 3x + 5y = 15 2.) 2x ­ 4y = 2 3.)18x + 12y = 24

Mathematics
1 answer:
DiKsa [7]3 years ago
4 0

1). Answer x= -5/3 y + 5

2.)  x=2y+1

3.)x=−2/3y+4/3

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For what value of x is the equation 6x-15=x+85
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Step-by-step explanation:

please give me brainlest

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andrew-mc [135]

For the reaction,

CO(g) + H_2 O(g) = CO_2(g) + H_2(g)

Initial concentration:

<u>0.1M</u> , <u>0.1M</u> , <u>0</u> , <u>0</u>

Let 'x' mole per litre of each of theproduct be formed.

At equilibrium:

<u>(0.1 – x)M</u> , <u>(0.1 – x)M</u> , <u>xM</u> , <u>xM</u>

where x is the amount of Carbon dioxide and Hydrogen, at equilibrium.

Hence, equilibrium constant can be written as,

K_c = \frac{x²}{(0.1 – x)²} = 4.24

→ x² = 4.24 (0.01 + x² – 0.2x)

→ x² = 0.0424 + 4.24 x² - 0.848x

→ 3.24x² - 0.848x + 0.0424 = 0

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(for quadratic equation ax² + bx+c=0)

x =  \frac{( - b \: ± \: \sqrt{ {b}^{2} - 4ac) } }{2a}

=  > x =  \frac{ - ( - 0.848 \: ± \:  \sqrt{( - 0.848)^{2} - 4(3.24)(0.0424) } }{2 \times 3.24}

=  > x =  \frac{ - 0.848±0.4118}{6.48}

x_1 =  \frac{0.848 - 0.4118}{6.48} = 0.067

x_2 =  \frac{0.848 + 0.4118}{6.48} = 0.194

Here, the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.

∴ The equilibrium concentrations are :-

[CO_2] [H_2] = x = 0.067M

[CO] [H_2 O] = 0.1 - 0.067 = 0.033M

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