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Masja [62]
3 years ago
15

The period 4 alkaline-earth metal has the name_______ . An isotope of this element having 20 neutrons has amass numberof _______

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
6 0

Answer:

The period 4 alkaline-earth metal has the name Calcium. An isotope of this element having 20 neutrons has a mass number of 40.

Explanation:

Looking at the periodic table 4th row for the alkaline-earth metal, we can find Calcium which has an atomic number of 20 (or protons number, or z). The mass number is given by the sum of protons and neutrons. If the number of protons in this case is 20 and the number of neutrons is also 20, the mass number A = 40.

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Which of the following best describes solids?
Angelina_Jolie [31]
In my opinion I would pick D. 

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But think of a block of ice. You can break it, and crush it. It cannot be formed into any other shape. If you did break it, it only would make more ice fragments. And it's next phase is a liquid; if it were to be melted.
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3 years ago
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56. What would be the valence electrons of 14Si, 16S, 32Ge respectively?
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Answer: D. 4,6,4

Explanation:

The valence electron is the electron found in the outermost shell of an atom, and participates in bond formation.

For the following element, the valence electron can be noticed when we write the electronic configuration of each element.

Silicon =  1s2 2s2 2P6 3s2 3p2

Sulphur = 1s2 2s2 2P6 3s2 3p4

Germanium =  1s2 2s2 2P6 3s2 3p6 3d10 4s2 4p2

8 0
3 years ago
What was observed when positively charged particles were radiated onto a gold atom during Rutherford's experiment because of thi
trapecia [35]

Answer:

He was happy:)

Explanation:

3 0
3 years ago
Is the following a physical or chemical change?
tester [92]

Answer:

chemical change

Explanation:

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6 0
2 years ago
One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
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