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3 years ago

The radioactive isotope Uranium-235 has a difference of ________________________ in comparison to Uranium-238.

Chemistry

1 answer:

Paraphin [41]3 years ago

8 0

Answer:

A) 3 less Neutrons

Explanation:

Uranium - 235 and Uranium - 238 have the same amount of protons because they are the same element: 92 protons.

However, U - 235 has 143 neutrons while U - 238 has 146 neutrons. U - 238 has a mass of 3 neutrons more.

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dezoksy [38]

Answer:

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ii. A bee sting - Toothpaste

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A bee sting is acidic so an alkaline substance such as toothpaste should neutralize it as well.

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2 years ago

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2 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

atmospheric pressure at elevations of 8000 feet averages about 0.72 atmospheres. Would a cabin pressurized at 500 mm hg meet fed

NeTakaya

Answer:

Yes, but it must be kept at that value and do not let it to decrease more.

Explanation:

Hello.

In this case, in order to substantiate whether the cabin meet the federal standards, we need to convert the 500 mmHg to atm and compare the result with 0.72 atm by knowing that 1 atm equals 760 mmHg:

$500mmHg*\frac{1atm}{760mmHg} \\\\=0.66atm$

Thus, since 0.66 atm is 0.06 atm away from the federal standard we can infer that it may meet the federal standard, however, it would not be recommended to let the pressure decrease more than that.

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3 years ago