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Mekhanik [1.2K]
3 years ago
9

the men's world record for swimming 1500.0 m in a long course pool (as of 2007) is 14 min 34.56 s. At this rate, how many second

s would it take to swim 0.150 miles?
Chemistry
1 answer:
Lyrx [107]3 years ago
5 0

140 s. It would take 140 s to swim 0.150 mi .

<em>Step 1</em>. Convert the <em>time to seconds</em>

Time = 14 min × (60 s/1 min) + 34.56 s = 840 s + 34.56 s = 874.56 s

<em>Step 2</em>. Convert <em>miles to metres </em>

Distance = 0.150 mi × (1609.3 m/1 mi) = 241.4 m

<em>Step 3.</em> Calculate the <em>time to swim 241.4 m</em>

Time = 241.4 m × (874.56 s/1500 m) = 140 s

(<em>As of 2012, the men’s freestyle record for 1500 m was 14:31.02</em>.)

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Find the number of Grams
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8 0
2 years ago
Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
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0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0
3 years ago
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