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elena-14-01-66 [18.8K]
1 year ago
14

A car uses 12.5 L of gasoline to travel a distance of 275 km. Convert this into units of miles per gallon (mi/gal).

Chemistry
1 answer:
Natasha2012 [34]1 year ago
4 0

Answer:

The mileage of the car is 51.75 mil/gal.

Explanation:

Distance covered by the car = 275 km = \frac{275}{1.609} mile

1 Mile = 1.609 kilometer

Volume of gasoline used by the car = 12.5 L =\frac{12.5}{3.785} gal

1 gallon = 3.785 Liter

Mileage of the car :

= \frac{\frac{275}{1.609} mile}{\frac{12.5}{3.785} gal}

=51.75 mil/gal

The mileage of the car is 51.75 mil/gal.

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Draw the diazonium cation formed when cytosine reacts with nano2 in the presence of hcl.
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Answer in the Word document below.
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2 years ago
Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample ha
Stells [14]

Answer:

296.1 day.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).

<em>∴ kt = lna/(a-x)</em>

(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>

5 0
2 years ago
Find the molarity of 750 ml solution containing 346 g of potassium nitrate
Zinaida [17]
Given mass of KNO₃=346g
Molar mass of KNO₃=(39.098)+(14)+(15.99*3)=101.068gmol⁻¹
Volume of Solution=750ml=0.75dm³

Molarity=(mass of solute/molar mass of solute)*(1/volume of sol. in dm³)
            =(346/101.068)*(1/0.75)
            =4.56 mol dm⁻³
5 0
2 years ago
A sample of a compound of Cl and O reacts with an excess of H2 to give 0.233g of HCl and 0.403g of H2O. Determine the empirical
Vsevolod [243]

Answer:

Cl₂O₇

Explanation:

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Moles of HCl and moles of H₂O are:

HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl

H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O

As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:

6.39x10⁻³ mol Cl

2.236x10⁻² mol O

Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is <em>1</em>, Oxygen will be:

2.236x10⁻² mol / 6.39x10⁻³ = <em>3.5</em>

As empirical formula must be given in natural numbers, the empirical formula is:

<em>Cl₂O₇</em>

<em></em>

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2 years ago
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