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3 years ago

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The intersection angle of a 3 degree curve is 45.2 degrees. What is the length of the curve? of Select one: O a. 455 m O b.573 m

C. 452 m O d. 25.9 km

1 answer:

ValentinkaMS [17]3 years ago

7 0

Explanation:

Relation between length of a curve and angle is as follows.

l = $R \times \Theta$

where, R = radius of curve

$\Theta$ = angle in radians

Also, l = $R \times \Theta \times \frac{\pi}{180}$ .......... (1)

If curve has a degree of curvature $D_{a}$ for standard length s, then

R = $\frac{s}{D_{a}} \times \frac{180}{\pi}$ ........... (2)

Now, substitute the value of R from equation (2) into equation (1) as follows.

l = $\frac{s \times \Theta}{D_{a}}$

If s = 30 m, then calculate the value of l as follows.

l = $\frac{s \times \Theta}{D_{a}}$

= $30 \times \frac{45.2}{3}$

= 452 m

thus, we can conclude that the length of the curve is 452 m.

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<u>Explanation:</u>

To calculate the molarity, we use the equation:

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Moles hydrochloric acid solution = 0.060 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.060}{1L}\\\\\text{Molarity of HCl}=0.060M$

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$C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-$

<u>Initial:</u> 0.24 0.060 0.31

<u>Final:</u> 0.18 - 0.37

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.2

$[C_6H_5COO^-]=0.18M$

$[C_6H_5COOH]=0.37M$

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Putting values in equation 1, we get:

$pH=4.2+\log(\frac{0.18}{0.37})\\\\pH=3.89$

To calculate the hydronium ion concentration in the solution, we use the equation:

$pH=-\log[H_3O^+]$

pH = 3.89

Putting values in above equation, we get:

$3.89=-\log[H_3O^+]$

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The product when an unsaturated hydrocarbon undergoes complete addition would be a saturated molecule. These saturated hydrocarbons becomes Alkanes where these molecules are only single-bonded in contrary to unsaturated hydrocarbons which have multiple bonds between atoms.

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Ideal gas law: PV = nRT

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V=10L
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Sub in:
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Mass of ozone( $O_{3}$ ) is calculated with the help of mass of $NO_{2}$ using stoichiometry.

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Step 2 : Find the moles of $O_{3}$ using moles of $NO_{2}$ .

Moles of $O_{3}$ is calculated by using moles of $NO_{2}$ with the help of mole ratio.

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From the balanced chemical equation we can see that coefficient of $NO_{2}$ is 1 and coefficient of O3 is 1 , so mole ratio of $O_{3}$ to $NO_{2}$ is 1:1

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Step 3 : Convert moles of $O_{3}$ to grams of $O_{3}$

Grams = Moles X Molar mass

Molar mass of $O_{3}$ = 48.0 g/mol

Grams = $(0.043 mol O_{3})\times (\frac{48 g O_{3}}{1 mol O_{3}})$

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Note : The above three steps can also be done using a single step setup.

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