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kupik [55]
3 years ago
11

Rolls of foil are 304 mmmm wide and 0.014 mmmm thick. (The density of foil is 2.7 g/cm3g/cm3 .) What maximum length of foil can

be made from 1.06 kgkg of foil?
Chemistry
1 answer:
ratelena [41]3 years ago
3 0

Answer:

92.2 m

Explanation:

Given that:=

The breadth = 304 mm

Height = 0.014 mm

Let Length = x mm

Volume = Length\times breadth\times height

Thus,

Volume = 304\times 0.014\times x\ mm^3=4.256x\ mm^3

Also, 1 mm³ = 0.001 cm³

So, volume = 0.004256 cm³

Given that density = 2.7 g/cm³

Mass = 1.06 kg = 1060 g

So,

Volume=\frac{Mass}{Density}=\frac{1060}{2.7}\ cm^3=392.59\ cm^3

So,

0.004256*x = 392.59

x = 92243.89 mm

Length of foil = 92243.89 mm = 92.2 m

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A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency
saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

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E = hf

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3 years ago
What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su
anastassius [24]

Answer:

V=27992L=28.00m^3

Explanation:

Hello,

In this case, the combustion of methane is shown below:

CH_4+2O_2\rightarrow CO_2+2H_2O

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3

Best regards.

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Answer:

Explanation:

8.25 atm

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