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4 years ago

Disadvantages of continued polymers

Chemistry

1 answer:

saw5 [17]4 years ago

4 0

Answer:

Cannot withstand very high temperature as all plastics melt down very soon as compared to metals

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The distance from the surface of the earth to the moon is approximately 3.8 * 10 to the 5th km. How many days would a distance l

Alla [95]

Answer:

82 days would take a distance like that, if we drive at 120 mi/h

Explanation:

Distance from Earth to moon is 3.8×10⁵ km

You drive at a speed of 120 mi/h

Let's convert the 120 mi/h to km/h

1 mi = 1.61 km

Then 120 mi . 1.61 km / 1mi = 193.2 km

New velocity is 193.2 km. Therefore,

193.2 km are driven in 1 hour

3.8×10⁵ km woul be driven in (3.8×10⁵ . 1)/ 193.2 = 1966.8 hours

1967 hours . 1 day / 24 hours = 82 days

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4 years ago

Give me an atom with the following characteristics:

guajiro [1.7K]

Explanation:

Lanthanide series= E4

Boron=Si

Chalogen=O

Alkaline Earth metal =M9

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3 years ago

When an electron move from a high energy level to a low energy, it ______ energy.

hjlf

Answer:

released often as light.

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3 years ago

Apple juice can be classified as which of the following?

HACTEHA [7]

Answer:

mixture

Explanation:

5 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Liquid nitrogen trichloride is heated in a 3.00−L closed reaction vessel until it decomposes comple

iris [78.8K]

Answer:

$p_{Cl_2}=629.25mmHg\\p_{N_2}=209.75mmHg$

Explanation:

Hello,

At first, consider the chemical reaction:

$2NCl_3-->3Cl_2+N_2$

Now, considering the ideal gas equation, we compute the total moles:

$PV=nRT\\n=\frac{PV}{RT} \\n=\frac{839mmHg*\frac{1atm}{760mmHg}*3L }{0.082\frac{atm*L}{mol*K}*359K }\\ n=0.1125mol$

Then, taking into account that the total moles are stoichiometrically handed out by the half (0.05625mol), one can say that:

$n_{Cl_2}=0.05625molNCl_3*\frac{3mol Cl_2}{2molNCl_3} =0.084375molCl_2\\n_{N_2}=0.05625molNCl_3*\frac{1mol N_2}{2molNCl_3} =0.028125molN_2$

Thus, the molar fractions are:

$x_{Cl_2}=\frac{0.084375}{0.1125} =0.75\\x_{N_2}=\frac{0.028125}{0.1125} =0.25$

Finally, the partial pressures are:

$p_{Cl_2}=x_{Cl_2}P=0.75*839mmHg=629.25mmHg\\p_{N_2}=x_{N_2}P=0.25*839mmHg=209.75mmHg$

Best regards.

8 0

3 years ago