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2 years ago

A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?

Chemistry

1 answer:

Tanzania [10]2 years ago

3 0

<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume = 0.50 cm × 10.0 cm × 2.0 cm

= 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

= 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

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What is the sequence of energy transformations that occur in a nuclear reactor? nuclear energy Right arrow. Mechanical energy Ri

Mademuasel [1]

Answer:

NUCLEAR ENERGY -----> MECHANICAL ENERGY -------> THERMAL ENERGY --------> ELECTRICAL ENERGY

Explanation:

In nuclear reactor, various energy transformations occur in order to generate electricity. Nuclear reactor converts the energy released from nuclear fission and the heat generated is removed from the reactor by a cooling system where steam is generated. The steam then drives a turbine which powers a generator to produce electricity.

A nuclear reactor is hence an equipment where nuclear chain reactions occur and control can be obtained. The nuclear reactor uses mostly uranium-235 and Plutonium-239. When these radioactive substances absorbs neutrons, they undergo nuclear fission causing the nucleus to split into two or more smaller compounds with the release of kinetic energy a form of mechanical energy, gamma radiations and others.The kinetic energy is then harnessed in the equipment as heat (thermal energy) which is received by a cooling system and steam is generated. The steam can then power the generator from which electricity is obtained (electrical energy).

So therefore, in a nuclear reactor, the nuclear energy is transformed to mechanical energy and then thermal energy which powers the generation of the electrical energy.

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2 years ago

The complex ion, [ni(nh3)6] 2+, has a maximum absorption near 580 nm. calculate the crystal field splitting energy (in kj/mol) f

Wewaii [24]

In given data:
maximum absorption wavelength λ = 580 nm = 580 x 10⁻⁹ m
write the equation to find the crystal field splitting energy:
E = hC / λ
Here, E is the crystal field splitting energy, h = 6.63 x 10⁻³⁴ J.sec is Planck's constant and C = 3 x 10⁸ m/sec is speed of light.
substitute in the equation above:
E = (6.64 x 10⁻³⁴ x 3 x 10⁸) / (580 x 10⁻⁹) = 3.43 x 10⁻¹⁹J
This crystal field splitting energy is for 1 ion.
Number of atoms in one mole, NA = 6.023 x 10²³
to calculate the crystal field splitting energy for one mole:
E(total) = E x NA
= (3.43 x 10⁻¹⁹) x (6.023 x 10²³) = 206 kJ/ mole

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2 years ago

Which of the following statements is true about energy quantization at the atomic level? Electrons in the outermost orbits are t

MAVERICK [17]

<h2>Answer:</h2>

The correct answer is option C which is, "Electrons in the orbit closest to the nucleus have the least amount of energy".

<h3>Explanation:</h3>

There are different orbitals around the nucleus on which the electrons moves around the nucleus.
These orbitals have a specific energy, due to which they are known as energy levels.
The energy level near to the nucleus has least amount of the energy and the energy of the orbitals increase as the distance of the orbitals increase to the nucleus.

8 0

2 years ago

Which electron has a higher probability of being found at 4 Å of the nucleus, one in a 2s orbital or one in a 2p orbital?

Lena [83]

Answer:

one in a 2s orbital

Explanation:

Because of the peak near the nucleus in the 2s curve there is a higher probability of finding a 2s within 4 Å of the nucleus. In a multi-electron atom an electron in a 2s orbital will have a lower energy than one in a 2p orbital

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2 years ago

You wish to make a 0.339 m h2so4 solution from a stock solution of 12.0 m h2so4. how much concentrated acid must you add to obta

Alex_Xolod [135]

When preparing diluted solutions from concentrated solutions , we can use the following equation;
c1v1 =c2v2
Where c1 and v1 are the concentration and volume of the concentrated solution
c2 is the concentration of the diluted solution to be prepared
v2 is the volume of the diluted solution
Substituting the values;
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v1 = 2.825 mL needs to be taken from the stock solution

8 0

2 years ago