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3 years ago

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Which of the following must be done at the end of an experiment?a) Clean all laboratory equipment and return it to the proper lo

cation.b) Wipe down your lab station.c) Wash your hands with soap.d) All of the above.

1 answer:

andriy [413]3 years ago

8 0

The answer D All of the above

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Imagine that you have many U-235 atoms and you fire a neutron at one of them. What do you think will happen?

Damm [24]

The correct answer is Release of large amount of heat and radiation.

U-235, which makes up just over 0.7% of natural uranium, is relatively uncommon; uranium is nearly 100 times more frequent than silver.Because uranium-235, one of the radioactive metal's isotopes, is the only naturally occurring isotope capable of supporting a nuclear fission process, radioactive metal is special. Nuclear power facilities, as well as the nuclear reactors that power ships and submarines, can be powered by uranium that has been "enriched" to have a higher proportion of U-235. In nuclear weapons, it can also be applied. Examples of extremely unstable fission fragments produced by uranium-235 fission events caused by neutron absorption include the barium and krypton nuclei. These pieces nearly instantly split into barium-144 and krypton-89 by releasing three neutrons between them.

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2 years ago

Does temperature stay the same when heat energy increases?

son4ous [18]

Answer:

More energy is required to raise its temperature. Therefore, temperature does not stay the same when heat energy increases.

6 0

3 years ago

Read 2 more answers

What is another name for producer why are they called this

just olya [345]

Answer:

Autotrophs

Explanation:

They are responsible for the production of food. Root words "auto" for self and "troph" for food.

3 0

3 years ago

I have a buddy who recycles electronics, and isolates metals from the connector pins electrical boards. He isolates gold, for ex

Kisachek [45]

Answer:

Theoretical yield = 3.75g

Explanation:

Percent yield is defined as one hundred times the ratio between actual yield and theoretical yield. The expression is:

Percent yield = Actual Yield / Theoretical Yield * 100

In the problem, your actual yield was 3-00g.

Percent yield is 80.0%.

Solving for theoretical yield:

80% = 3.00g / Theoretical yield * 100

Theoretical yield = 3.00g / 80.0% * 100

<h3>Theoretical yield = 3.75g</h3>

3 0

3 years ago

For most solids at room temperature, the specific heat is determined by oscillations of the atom cores in the lattice (each osci

Cloud [144]

Answer:

The specific heat of copper is $C= 392 J/kg\cdot ^o K$

Explanation:

From the question we are told that

The amount of energy contributed by each oscillating lattice site is $E =3 kT$

The atomic mass of copper is $M = 63.6 g/mol$

The atomic mass of aluminum is $m_a = 27.0g/mol$

The specific heat of aluminum is $c_a = 900 J/kg-K$

The objective of this solution is to obtain the specific heat of copper

Now specific heat can be defined as the heat required to raise the temperature of 1 kg of a substance by $1 ^o K$

The general equation for specific heat is

$C = \frac{dU}{dT}$

Where $dT$ is the change in temperature

$dU$ is the change in internal energy

The internal energy is mathematically evaluated as

$U = 3nk_BT$

Where $k_B$ is the Boltzmann constant with a value of $1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K$

T is the room temperature

n is the number of atoms in a substance

Generally number of atoms in mass of an element can be obtained using the mathematical operation

$n = \frac{m}{M} * N_A$

Where $N_A$ is the Avogadro's number with a constant value of $6.022*10^{23} / mol$

M is the atomic mass of the element

m actual mass of the element

So the number of atoms in 1 kg of copper is evaluated as

$m = 1 kg = 1 kg * \frac{10000 g}{1kg } = 1000g$

The number of atom is

$n = \frac{1000}{63.6} * (6.0*0^{23})$

$= 9.46*10^{24} \ atoms$

Now substituting the equation for internal energy into the equation for specific heat

$C = \frac{d}{dT} (3 n k_B T)$

$=3nk_B$

Substituting values

$C = 3 (9.46*10^{24} )(1.38 *10^{-23})$

$C= 392 J/kg\cdot ^o K$

5 0

3 years ago