Evaluate 15 ÷ (-3) + 5|-2| ÷ (-5)

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andrew-mc [135]

For the reaction,

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

Initial concentration:

0.1M , 0.1M , 0 , 0

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

At equilibrium:

(0.1 – x)M , (0.1 – x)M , xM , xM

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

a = 3.24, b = -0.848, c = 0.0424

(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 :-

$[CO_2] [H_2] = x = 0.067M$

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

7 0

3 years ago

Steven has 5/6lb of jelly beans and 1/3lbs of gummy bears. How many more pounds of jelly beans than gummy bears does he have ?

GalinKa [24]

1/3=2/6

5/6-2/6= 3/6 or 1/2

He has 1/2 pound of more jelly beans than gummy bears.

☺☻♥Hope I helped!☻☺♥

Good Luck!♣

8 0

4 years ago

Read 2 more answers

Jan'ai was asked to determine the minimum for a function with zeros located at –1 and 5, which also has a y-intercept of (0, –25

sergey [27]

The first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

<h3>What are the coordinates of vertex for a quadratic function?</h3>

For a quadratic function of the form $y = ax^2 + bx + c$ , its vertex form is obtained as:

$y = ax^2 + bx + c\\y =a(x^2 + bx/a) + c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2 -(b/2a)^2 )+ c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2) -a \times (b/2a)^2 + c\\\\y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$

For the form $y = a(x-h)^2 + k$ , the vertex has coordinates (h, k)

Thus, for the obtained equation $y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$ , we get the coordinates of vertex as:

$h = -b/2a$ , $k = c - a\times(b/2a)^2$

Thus, the coordinates of vertex of $y = ax^2 + bx + c$ is:

$(h,k) = (-b/2a, c - a \times (b/2a)^2 )$

The missing steps of work of Jan'ai are:

Begin to write a function in factored form. $f(x) = a(x+1)(x-5)$
Substitute x = 0, y = f(x) = -25 to determine a. $-25 = a(0+1)(0-5)$
Simplify and solve to find a. $a=5$
Rewrite the function. $f(x) = 5(x+1)(x-5)$
Rewrite in standard form. $f(x) = 5x^2-20x-25$
Find the x-coordinate of the vertex. $x = -20/2(5) = -20/10; x = -2$
Find the y-coordinate of the vertex.

$y = 5x^2-20x-25$

$y = 5(-2)^2-20(-2)-25$

$y = 35$

so (-2,35) is the coordinate of the vertex, which denotes the minimum.

So, as we see, in the 5th step, Jan'ai had the quadratic function $f(x) = 5x^2-20x-25$ ,

Comparing this to $f(x) = ax^2 + bx + c$ , we get a = 5, b = -20, c = -25

The vertex's x-coordinate will be on -b/2a = -(-20)/ 2(5) = 20/10 = 2

But Jan'ai didn't putted that negative sign before b. in the 6th step.

Thus, the first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

Learn more about the vertex form of a quadratic equation here:

brainly.com/question/9912128

6 0

3 years ago

HELP PLS ASAP.I LOST SO MY POINTS

Leto [7]

X=6 because 6 x 6 x 6 = 216

7 0

3 years ago

Read 2 more answers

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ivolga24 [154]