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Katyanochek1 [597]
1 year ago
11

What is the yield of uranium from 2.50 kg U3O8?

Chemistry
1 answer:
mr_godi [17]1 year ago
8 0

Answer: Thus the yield of uranium from 2.50 kg U_3O_8 is 2.12 kg

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number (6.023\times 10^{23}) of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}

moles of U_3O_8=\frac{2.50\times 1000g}{842g/mol}=2.97mol    (1kg=1000g)

As 1 mole of U_3O_8 contains = 3 moles of U

2.97 mole of U_3O_8 contains = \frac{3}{1}\times 2.97=8.91moles moles of U

Mass of Uranium=moles\times {\text {Molar mass}}=8.91mol\times 238g/mol=2120g=2.12kg  

 ( 1kg=1000g)

Thus the yield of uranium from 2.50 kg U_3O_8 is 2.12 kg

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a solution of household vinegar is to be analyzed. A pipet is used to measure out 10 ml of the vinegar which is placed in a 250m
DerKrebs [107]

Answer:

0.0344 M

Explanation:

  • HC₂H₃O₂ (aq) + OH⁻ (aq) --> C₂H₃O₂⁻ (aq) + H₂O (l)

Because one mol of vinegar (acetic acid) reacts with one mol of NaOH, we can use the formula

  • C₁V₁=C₂V₂

Where C₁ and V₁ refer to the concentration and volume of vinegar, and C₂ and V₂ to those of NaOH. We're given V₁, C₂ and V₂; so we <em>solve for C₁</em>:

  • C₁ = C₂V₂ / V₁
  • C₁ = 0.0500 M * 16.7 mL / 25.0 mL
  • C₁ = 0.0344 M
6 0
2 years ago
What are the 3 critical components of an electromagnet and what purpose do they each serve?
WARRIOR [948]

Answer:

A source of electricity, a wire coil, and an iron core  

Explanation:

An electromagnet has three critical components:

1. A source of electricity

This is often a battery.

It generates the electric current that produces the magnetic field.  

2. A wire coil

The wire carries the electric current.

Stacking the wire into loops makes a stronger magnetic field.

The more loops in the coil, the stronger the field.

3. An iron core

An iron core greatly increases the strength of the magnetic field within it and at its ends.

4 0
2 years ago
En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello
MAXImum [283]

Answer:

a) m_{CH_4}=2630kg

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}

m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol

Con ello, los kilogramos de metano que cuestan 0,45 €:

44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0
2 years ago
The table above summarizes data given to a student to evaluate the type of change that took place when substance X was mixed wit
Tamiku [17]

The question is incomplete, the complete question is;

The table above summarizes data given to a student to evaluate the type of change that took place when substance X was mixed with water. The student claimed that the data did not provide enough evidence to determine whether a chemical or physical change took place and that additional tests were needed. Which of the following identifies the best way to gather evidence to support the type of change that occurred when water and Xwere mixed?

A. Measuring the melting point of the mixture of water and X

B. Adding another substance to the mixture of water and X to see whether a solid forms

C Measuring and comparing the masses of the water,  X, and the mixture of water and X

D Measuring the electrical conductivities of X and the mixture of water and X

Answer:

D Measuring the electrical conductivities of X and the mixture of water and X

Explanation:

Unfortunately, I am unable to reproduce the table here. However, from the table,  the temperature of the of the mixture of the solid X and water was 101.6°C. This is above the boiling point of water and way below the temperature of the solid X.

This goes a long way to suggest that there was some kind of interaction between the water and X which accounted for the observed temperature of the system of X in water.

The only way we can be able to confirm if X actually dissolved in water is to measure the conductivity of the water. dissolved solids increase the conductivity of water.

6 0
1 year ago
THESE QUESTIONS ARE ELECTRICIAN REALATED
Vladimir [108]

<span>1.      </span>A.

It is a technique typically engaged in electrical power and electronic devices, wherein the devices are run at less than their rated maximum power degeneracy, taking into account the case or body temperature, the ambient temperature and the type of cooling mechanism used.

<span>2.      </span>B.

In order to answer the question, you need to know the area of a circle.<span>
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Area = 3.14 * (100/2) ^ 2
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<span>3.      </span>A.

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RHW cale insulation can be used up to 167 degrees F.

<span>5.      </span>D.

This is a type of electrical connector used to fasten two or more low-voltage (or extra-low-voltage) electrical conductors.

<span>6.      </span>C. the ease with which a material allows electricity to move is called CONDUCTIVITY

<span>7.      </span>D.

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<span>8.      </span>D.

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<span>9.      </span>D.

A #12 copper conductor with an insulation factor of 90 degrees C is rated at 20 amps. A #12 aluminum conductor with an insulation rating of 90 degrees C is rated at 15 amps. These conductors ratings only applies to three conductors in a rated at 15 amps. These conductors’ ratings only applies to three conductors in a raceway. From 7 to 24 conductors in a raceway, both aluminum and copper conductor's ratings have to be reduced by .70, so 15 amps x .7 = 10.5 amps and 20 amps x .7 = 14 amps respectively.

<span>10.  </span>A.

<span>NEC 310.15(B)(1) and (2), table 310.15(B)(2)(a), and table 310.15(B)(17).
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</span>

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<span>14.  </span>D.

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<span>15.  </span>A.

<span>16.  </span>D.

<span>17.  </span>C.

Table 310-16 NEC

<span>18.  </span>A <span>
MCM means 1000's (M) of (C)ircular (M)ils</span>

<span>19.  </span>D.

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4 0
2 years ago
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