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3 years ago

Why are electromagnets used in metal scrap yards

Chemistry

2 answers:

Rudik [331]3 years ago

7 0

Theyre temporary magnets, so they can be turned on and off.

Alenkinab [10]3 years ago

5 0

Answer:c

Explanation: they can be turned on and off

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For the reaction 2A(g) â B(g), the equilibrium constant is Kp = 0.76. A reaction mixture initially contains 4.0 atm of gas (PA =

timofeeve [1]

Answer: (a) The reaction mixture will proceed toward products.

Explanation:

Equilibrium constant is defined as the ratio of pressure of products to the pressure of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

$2A(g)\rightleftharpoons B(g)$

The expression for $Q_p$ is written as:

$Q_p=\frac{p_B}{(p_A)^2}$

$Q_p=\frac{(2.0)}{(2.0)^2}$

$Q_p=0.5$

$K_p=0.76$

Thus as $K_p>Q_p$ , the reaction will shift towards the right i.e. towards the product side.

7 0

2 years ago

Students are measuring the length of time it takes for soda to go flat at three different temperatures. Which type of graph shou

Akimi4 [234]

The answer would be a bar graph

3 0

3 years ago

A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is

Romashka [77]

Answer:

$\large \boxed{\text{B.) 2.8 atm}}$

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 1520 Torr; T₁ = 27 °C

p₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\$

(c) Convert the pressure to atmospheres

$p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}$

7 0

2 years ago

During the apollo missions,an astronaut took a golf ball to the moon. He placed it on the moon’s surface and hit the ball with a

aleksandr82 [10.1K]

Answer:

b.)

Explanation:

Because you would have a different weight on the moon because of it's low gravity

3 0

3 years ago

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

8 0

3 years ago