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3 years ago

What is the x-coordinate of the function below when the y-coordinate is 0?(0,5) (2,-5)

Chemistry

1 answer:

maria [59]3 years ago

4 0

The equation of the function is assumed to be a straight line is of the form,

$y= mx+c$ , where $m,c$

are the gradient and the y intercept respectively.

The gradient is computed using the formula for the gradient, i.e

$m=\frac{y_2-y_1}{x_2-x_1} = \frac{-5-5}{2-0} = -5$ .

The equation of the line is,

$y= -5x+c\\=>5=-5(0)+c\\=> c= 5$

The equation of the function is ,

$y = -5x+5.$ .

The value of the x-coordinate when the y coordinate is 0 is found as shown below.

$0=-5x+5\\=>x=1\\$ .

so the x-coordinate is 1.

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Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc

Ber [7]

Answer:

The balanced equation for this reaction will be

$CH4 + 4F2$ → $CF4 + 4HF$

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

1 mole of CH4 → 4 mole of 4 mole of fluorine

0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction

but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

4 moles of fluorine → 1 mole of CF4

0.56 mole → $\frac{1}{4} * 0.56$ = 0.14mole of CF4

4 moles of fluorine → 4 moles of HF

0.56 mole of fluorine → 0.56 mole of HF

now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.

8 0

2 years ago

Given a gas whose temperature is 418 K at a pressure of 56.0 kPa. What is the pressure of the gas if its Temperature changes to

Rainbow [258]

Answer: P₂=0.44 atm

Explanation:

For this problem, we are dealing with temperature and pressure. We will need to use Gay-Lussac's Law.

Gay-Lussac's Law: $\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }$

First, let's do some conversions. Anytime we deal with the Ideal Gas Law and the different laws, we need to make sure our temperature is in Kelvins. Since T₂ is 64°C, we must change it to K.

64+273K=337K

Now, it may be uncomfortable to use kPa instead of atm, so let's convert kPa to atm.

$56.0kPa*\frac{1000Pa}{1kPa} *\frac{atm}{101325Pa} =0.55atm$