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lisov135 [29]
2 years ago
14

How many atoms and ions does nickel oxide have?

Chemistry
1 answer:
Alex73 [517]2 years ago
4 0

Answer: 1 atom

Explanation:

The loss of the remaining valence electron results in an ion with a +2 charge. The proper way of noting the charges on these ions is to use the systematic name for each ion, nickel (I) for the +1 ion and nickel (II) for the +2 ion.

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Any help with atomic radius, ionization & electronegativity? See picture please! I dont really understand any of the three.​
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a. Ra
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19)
a. C
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a. S
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Sorry wasn’t sure about some of them in #18
Hope the notes are helpful.

8 0
2 years ago
1. Calculate the percent oxygen in H2SO4.
Mashutka [201]
I believe the answer is 65.254%
6 0
2 years ago
Iron(II) sulfide has a primitive cubic unit cell with sulfide ions at the lattice points.
masya89 [10]

We have that the  the density of FeS  is mathematically given as

  •  \phi=2.56h/cm^3

From the question we are told

Iron(II) sulfide has a primitive <em>cubic</em> unit cell with <em>sulfide</em> ions at the <em>lattice points.</em>

The ionic radii of iron(II) ions and sulfide ions are 88 pm and 184 pm, respectively.

What is the density of FeS (in g/cm3)?

<h3>Density</h3>

Generally the equation for the Velocity  is mathematically given as

V_c=a^3=(2\pi Fe^{2+}+2\pi S^{2-})^3\\\\Therefore\\\\V=a^3(2\pi*0.088+2\pi 0.184)^3\\\\V=16.98*10^{-23}\\\\Therefore\\\\\phi=n\frac{PFeion+PSion}{VNa}\\\\\phi=3*\frac{55.85+32}{16.9*10^{-23}*6.023*10^{23}}

  • \phi=2.56h/cm^3
  • \phi=2.56h/cm^3

For more information on ionic radii visit

brainly.com/question/13981855

4 0
2 years ago
Serial dilution problem: Six test tubes are placed in a rack. To each tube add 4 mL of saline solution. Now to the first tube ad
ArbitrLikvidat [17]

Answer:

The dilution factor of protein in tube # 4 is 125. Molar concentration is 0.0088 M protein

Explanation:

The dilution factor indicates how many times is more concentrated a main solution in relationship with a diluted solution. In this case, the main solution is in tube #1. For calculating the dilution factor and molar concentration in tube #4 we need the main solution concentration which comes from next equation:

Initial volume * initial concentration = final volume * final concentration

0.5 mL * 10M = 5mL * final concentration

1.1 M = final concentration = main solution concentration

Applying the same equation for remain tubes we have 0.22 M for tube #2, 0.044 M for tube # 4 and 0.0088 for tube # 4.

Dilution factor = Main solution concentration/tube 4 concentration

Dilution factor = 1.1/0.0088 = 125

I hope my answer helps you

3 0
3 years ago
What is the concentration of a solution made with 0.150 moles of KOH in 400.0 mL of solution?
otez555 [7]

Answer:

concentration = \frac{0.15}{0.4}=0.375 mol/L

Explanation:

Concentration: i is defined as the mole per litre.

concentration = \frac{mole}{volume in L}

mole=0.15

volume=400 ml=0.4 litre

concentration = \frac{0.15}{0.4}=0.375 mol/L

6 0
2 years ago
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