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Salsk061 [2.6K]
3 years ago
15

The nucleus is:

Chemistry
1 answer:
julsineya [31]3 years ago
8 0
The Nucleus is the center of the atom. What specifies the atom number is the amount of protons it has so it wouldn't  be B.  The number of nucleons (both protons and neutrons) in the nucleus is the atom's <span>mass number.</span>
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If 1 egg and 1/3 cup of oil are needed for each bag of brownie mix, how many bags of brownie mix do you need if you want to use
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The answer is 3 bags total
5 0
3 years ago
Give the word and chemical equations (including state symbols) for the reaction of potassium and water
zepelin [54]

Answer:

Zinc is Zn and Magnesium is Mg

7 0
3 years ago
Which gases are responsible for global warming?
elena-s [515]

Answer:  

Greenhouse gas Chemical formula Global Warming Potential, 100-year time horizon

Carbon Dioxide

Methane

Nitrous Oxide

Chlorofluorocarbon

Explanation:

8 0
3 years ago
List the protons, neutrons, and electrons for Ge2+<br> Please hurry
Nady [450]

Answer:

32, 30 and 41

Explanation:

The problem here is to find the number of:

    Protons, neutrons and electrons in Ge²⁺

In this ion,

   We must understand that for a net positive charge to remain on an atom, the number of protons must be greater than the number of electrons.

Ge is Germanium with atomic number of 32;

So the number of protons is 32

Since the atom has lost two electrons;

  Number of electrons now is 32  - 2 = 30

Number of neutrons is 41 from the periodic table.

7 0
2 years ago
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
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