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larisa [96]
3 years ago
7

Which element has similar properties to Berylium? Explain

Chemistry
1 answer:
GaryK [48]3 years ago
7 0
This has multiple answers can you be more specific
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Air is composed of 79%N^2, 21% O^2. if the air pressure is 101.3 kPa, what are the pressures of N^2 and O^2
pychu [463]
According to Dalton's law of partial pressures, a mixture of gases inflict pressure in a direct proportion to their volume/mass etc. Therefore, since 79 percent of the air is composed of nitrogen and 21 percent of Oxygen, we multiply 101.3 kPa by 0.79 = 80.027 kPa by Nitrogen, 101.3 kPa by 0.21 = 21.273 kPa by Oxygen.
5 0
3 years ago
Computers have the mineral _______________ in their motherboard.
lana66690 [7]

Answer:

silica

Explanation:

6 0
2 years ago
Read 2 more answers
What is the theoretical yield for 3.14 g of magnesium, and .367 g of magnesium?
Dennis_Churaev [7]

Answer:

Explanation:

i dont know

Magnesium is a chemical element with the symbol Mg and atomic number 12. It is a shiny gray solid which bears a close physical resemblance to the other five elements in the second column of the periodic .

Symbol: Mg

Atomic mass: 24.305 u

Electron configuration: [Ne] 3s2

Atomic number: 12

Melting point: 650 °C

Oxidation number: 2

6 0
3 years ago
In the following equation, ______ is being oxidized and ______ is being reduced.
Mashcka [7]

oxidation \: number \: of \: oxygen =  \\ before \: rxn =  - 2 \\ after \: rxn =  - 2

oxidation \: number \: of \: hydrogen = \\ before \: rxn =  + 1 \\ after \: rxn =  \\ 2x - 2 = 0 \\ x =  + 1

oxidation \: number \: of \: carbon =  \\ before \: rxn =  \\ x  - 6 =  - 2 \\ x = 4 \\ after \: rxn =  \\ x - 4 = 0 \\ x = 4

<h2>Option A</h2>

oxidation \: numbers \: remain \: constant \\ so \: none \:a re \: undergoing \: oxidation \: \\ nor \: reduction \:

6 0
1 year ago
A sodium ion, Na+, with a charge of 1.6×10−19C and a chloride ion, Cl− , with charge of −1.6×10−19C, are separated by a distance
sasho [114]

Answer:

W\geq 2.1x10^{-19}J

Explanation:

Due to Coulomb´s law electric force can be described by the formula F=K\frac{q_{1}.q_{2}}{r^{2}}, where K is the Coulomb´s constant (9x10^{9} N\frac{m^{2} }{C^{2} }), q_{1}= Charge 1 (Na+ in this case), q_{2} is the charge 2 (Cl-) and r is the distance between both charges.

Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is W=W_{f} -W_{i}.

so we have W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]

Given that ri= 1.1nm= 1.1x10^{-9}m and rf= infinite distance

W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J

6 0
3 years ago
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