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3 years ago

How many minutes will it take for a car traveling 74 miles per hour to cover 6.50 kilometers?

Chemistry

1 answer:

Dvinal [7]3 years ago

4 0

Answer:

3.27 minutes

Explanation:

To solve this problem we will first convert miles/hour to km/min. For this purpose we will require following reference conversions as;

1 Mile = 1.60934 Km

And,

1 Hour = 60 min

Therefore,

As,

1 Mile = 1.60934 Km

Then,

74 Miles = X Km

Solving for X,

X = 74 miles × 1.60934 Km ÷ 1 mile

X = 119.10 Km

Therefore,

119.10 Km were travelled in = 60 min

So,

6.50 Km will be travelled in = X min

Solving for X,

X = 60 min × 6.50 Km ÷ 119.10 Km

X = 3.27 minutes

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A student needs to prepare a stock solution for the lab - 500.0 ml of 0.750 m nitric acid. The student is provided with concentr

dlinn [17]

Molarity of concentrated nitric acid = 15.9 M

Volume of the stock solution to be prepared = 500.0 mL

Concentration of the stock that is to be prepared = 0.750 M

Calculating moles from molarity and volume of stock:

$500.0mL*\frac{1L}{1000mL}*0.750\frac{mol}{L} =0.375 mol HNO_{3}$

Calculating volume of concentrated nitric acid to be taken for the preparation of stock solution:

$0.375mol*\frac{1L}{15.9mol} =0.0236 L$

Converting L to mL:

$0.0236L*\frac{1000mL}{1L}$ = 23.6 mL

Volume of distilled water to be added to 23.6 mL of 15.9 M nitric acid to get the given concentration = 5000.0mL-23.6mL=976.4 mL

Therefore, 976.4 mL distilled water is to be added to 23.6 mL of 15.9 M nitric acid solution to prepare 500.0 mL of 0.750M nitric acid.

6 0

4 years ago

Click on the graph to show the correct relationship between altitude and boiling point

spin [16.1K]

Explanation:

This graph shows that an increase in altitude causes a relative decrease in boiling point temperatures. The two variables, it can be said, are inversely proportional.

This is because, with an increase in altitude, there is a proportionate decrease in air pressure. This means the vapor pressure of the fluid becomes strong enough, at a relatively lower temperature, to overcome the air pressure and for the liquid to boil.

This is why water can even boil without inputting heat into it but rather by just reducing the ambient air pressure.

8 0

4 years ago

Read 2 more answers

A nonpolar solute will dissolve in a polar solvent if the temperature is increased,

ANTONII [103]

Answer:

Explanation:

False

Because " like dissolve like " polar solute dissolve in polar solvent and vice versa

8 0

3 years ago

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0

3 years ago

A sugar solution is 15% pure. How

Arada [10]

Answer:

Amount of sugar in solution = 82.5 gram

Explanation:

Given:

Sugar in solution percentage = 15%

Total solution = 550 grams

Find:

Amount of sugar in solution

Computation:

Amount of sugar in solution = Sugar in solution percentage x Total solution

Amount of sugar in solution = 15% X 550

Amount of sugar in solution = 82.5 gram

8 0

3 years ago