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lys-0071 [83]
3 years ago
6

How long in hours would it take a runner to finish an 11 km race if she is running at a pace of 5.7 mi/hr?

Chemistry
2 answers:
Alenkinab [10]3 years ago
6 0

It would take 1.16 h for the runner to finish the race .

<em>Step 1. </em>Convert <em>kilometres to miles</em>.

11 km × (0.621 mi/1 km) = 6.83 mi

<em>Step 2</em>. Calculate the <em>time</em>.

Time = 6.83 mi × (1 h/5.7 mi) = 1.16 h

Oliga [24]3 years ago
5 0

Answer:  1.199 hour runner will require to cover the 11 km .

Explanation:

Distance covered by  runner = 11 km = 11 × 0.6213 miles = 6.8343 miles

1 km = 0.6213 mile

Speed of the runner = 5.7 mile/hour

speed=\frac{distance}{time}

5.7 mile/hour=\frac{6.8343 miles}{time}

time=\frac{6.8343 miles}{5.7 mile/hour}=1.199 hour

1.199 hour runner will require to cover the 11 km .

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There are on average 43 g of sugar and 355 mL can of soda please calculate the molarity of sugar in the can of soda the molar ma
vfiekz [6]

Explanation:

Given :

Amount of solute - sucrose (C12H22O11) = 41 g

Amount of solvent -soda  = 355-mL

Molarity of the solution with respect to sucrose= ?

Molarity(M) is a unit of concentration measuring the number of moles of a solute per liter of solution. The SI unit of molarity is mol/L.

Formula to find the molarity of solution :

               Molarity =  

Amount of solvent is given in mL, let’s convert to L :

               1 L = 1000 mL

Therefore, 355 mL in L will be :

                   

               = 0.355 L

We have the amount of solute in g, let’s calculate the number of moles first :

       Number of moles (n) =  

Molar mass of C12H22O11 = 342.29 g/mol.

Therefore, n =  

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8 0
3 years ago
Mercury is the only metal that is a liquid at room temperature. When mercury vapor is inhaled, it is readily absorbed by the lun
Lapatulllka [165]

Answer:

P = 0.0166 mm Hg

Explanation:

To solve this question, we need to use the Clausius Clapeyron equation, which is a commonly used expression to calculate vapour pressure at a given temperature. We have the enthalpy of vaporization of the mercury, so, let's write the equation:

Clausius Clapeyron equation:

Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁)    (1)

Where:

R: universal constant of gases (8.314 J / K.mol)

P₂: Vapour pressure at 43°C (or 316 K)

P₁: Pressure of mercury at the boiling point (1 atm)

T₂: temperature at 43 °C

T₁: Boiling point of mercury (357 °C or 630 K)

As we are given the boiling point of the mercury, we can safely assume that the pressure at this point is 1 atm, becuase remember that when a sustance boils, is because it's internal pressure has reached the atmospherical pressure of 1 atm. With this clear, all we just need to do is solve for P₂. We are going to do this very slowly so you can understand the process. First let's replace the given data:

Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/316 - 1/630)

Ln P₂ = -7108.49 * (3.16x10⁻³ - 1.59x10⁻³)

Ln P₂ = -7108.49 * (1.51x10⁻³)

Ln P₂ = -10.7338

P₂ = 10⁽⁻¹⁰°⁷³³⁸⁾

P₂ = 2.18x10⁻⁵ atm

We can express this value in mm Hg and it will be:

P₂ = 2.18x10⁻⁵ * 760

<h2>P₂ = 0.0166 mm Hg</h2>

Hope this helps

8 0
2 years ago
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