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ale4655 [162]
3 years ago
13

By mistake, a quart of oil was dumped into a swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm

3. Assuming the oil spread into a uniform, thin layer over the water, how thick was the oil slick?
Chemistry
1 answer:
kondor19780726 [428]3 years ago
3 0

The oil slick thick = 1.256 x 10⁻⁴ cm

<h3>Further explanation</h3>

Volume is a derivative quantity derived from the length of the principal

The unit of volume can be expressed in liters or milliliters or cubic meters

The conversion is

1 cc = 1 cm3

1 dm = 1 Liter

1 L = 1.06 quart

<em>so for 1 quart = 0.943 L</em>

\tt 0.943~L=0.943\times 10^{-3}m^3

Volume of oil dumped = volume of swimming pool

\tt 0.943\times 10^{-3}~m^3=25\times 30\times h(h=thick)\\\\h=\dfrac{0.943\times 10^{-3}}{750~m^2}=1.257\times 10^{-6}~m=\boxed{\bold{1.256\times 10^{-4}~cm}}

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lesantik [10]

Answer:

\large \boxed {2.8 \times 10^{23}\text{ atoms Ar}}

Explanation:

\text{Atoms of Ar} = \text{0.47 mol Ar} \times \dfrac{ 6.022  \times 10^{23}\text{ atoms Ar}}{\text{1 mol Ar}}\\\\= \large \boxed {\mathbf{2.8 \times 10^{23}}\textbf{ atoms Ar}}

3 0
3 years ago
Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

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The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

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2 moles of butane releases heat = 5314.8 kJ

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3 0
3 years ago
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I hope this helps.
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