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garik1379 [7]
3 years ago
13

Using the formula for velocity shown below, what is the average velocity of a bicyclist who biked 20 miles north of where he beg

an for a duration of 4 hours?
Chemistry
1 answer:
Marat540 [252]3 years ago
3 0

8.1km/hr due North

Explanation:

Given parameters:

Displacement = 20 miles North

Time = 4hours

Unknown;

Velocity = ?

Solution:

Velocity is the speed in a particular direction. It is simply displacement divided by time. A change in position with time.

                Velocity = \frac{displacement}{time}

The unit is usually expressed in km/hr or m/s

We need to convert the given displacement to that in km:

              1 mile = 1.609km

             20miles = 1.609 x 20,   32.18km

Velocity  =   \frac{32.18}{4} = 8.1km/hr due North

Learn more:

Velocity brainly.com/question/10962624

#learnwithBrainly

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Afina-wow [57]

Answer:

An addition reaction

Step-by-step explanation:

In an addition reaction, two or more molecules come together to form a single product, for example,

C₂H₂ + 2Cl₂ ⟶ C₂H₂Cl₄

This reaction consists of two successive additions. The product of the first reaction becomes a reactant and adds a second molecule of Cl₂ to form C₂H₂Cl₄

    C₂H₂ + Cl₂ ⟶ <em>C₂H₂Cl₂ </em>

<em><u>C₂H₂Cl₂</u></em><u> + Cl₂ ⟶ C₂H₂Cl₄ </u>

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4 0
3 years ago
n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a ph
olasank [31]

Answer:

Molarity: 0.111M

% (w/w): 0.666

Explanation:

The reaction of NaOH with acetic acid (CH₃COOH) is:

NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O

<em>where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.</em>

As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:

0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:

4.169x10⁻³ moles of CH₃COOH.

Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:

4.169x10⁻³ moles of CH₃COOH / 0.03754L = <em>0.111M</em>

<em></em>

As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:

4.169x10⁻³ moles × (60g / mol) = <em>0.2501 g of acetic acid</em>

Now, assuming density of solution as 1.00g/mL, 37.54mL weights <em>37.54g</em>.

Thus, percent by weight is:

0.2501g CH₃COOH / 37.54g × 100 = <em>0.666% (w/w)</em>

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