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FinnZ [79.3K]
3 years ago
13

Metals are{blank}because light reflects off the electrons.

Chemistry
2 answers:
Snezhnost [94]3 years ago
5 0

Answer:

shiny

Explanation:

Talja [164]3 years ago
3 0

Answer:

shiny

Explanation:

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An ion of iron has an 26 protons, 30 neutrons, and 23 electrons. What are its atomic number, atomic mass, and net charge?
sleet_krkn [62]

Answer:

Atomic number 26,atomic mass 26+30=56 dalton and net charge is 3+

Explanation:

The total number of proton present in an atom is known as the atomic number of that atom.From that point of view the atomic number of iron ion is 26.

  The total number proton and neutrons present in the nucleus of an atom or ion is termed as atomic mass.From that point of view the the atomic mass of iron ion is 26+30=56 dalton

 According to the given question iron ion contain 3 more protons than electrons as a result the iron ion will contain 3 unit positive charge.

5 0
3 years ago
Hello <3 I just need some help with periodic table of elements.
garik1379 [7]

Answer:

11

Explanation:

An atomic number of 11 means this atom will have 11 protons. A mass number of 23 means 23 - 11 this atom will have 12 neutrons. Since this atom is neutral the positive protons must be equal to the negative electrons. This atom will have 11 electrons

8 0
2 years ago
Read 2 more answers
In a climatological sense, dryness is a function of both annual rainfall and ________.
Margarita [4]

Answer:

In a climatological sense, dryness is a function of both annual rainfall and evaporation

8 0
2 years ago
What’s the answers to these?
nata0808 [166]

Answer:

0bervation

Explanation:

cause it is

5 0
3 years ago
Consider a solution containing .100 M fluoride ions and .126M hydrogen fluoride. The concentration of fluoride ions after the ad
S_A_V [24]
Given:

Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M

Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution

Assume: 50:50 ratio of fluoride ions and HF

12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-

5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles

Assume: Volume additive

Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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3 0
3 years ago
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