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NemiM [27]
3 years ago
6

What atom has 8 protons and 10 electrons?

Chemistry
2 answers:
Lorico [155]3 years ago
8 0

Answer:

In your case, you have 8 protons and 10 electrons. This means that the atom will carry a total 2 xx (1-) = 2- net charge. Therefore, you are dealing with an anion that carries a 2- net charge, i.e. it has two more electrons than it does protons.

maria [59]3 years ago
8 0
The atom that has 8 protons and 10 electrons is Oxygen ion
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When hno2 dissociates in water, which atom is dontaing electrons, allowing water to act as a lewis base?
miss Akunina [59]

Answer to this is O-atom.

Explanation: The Bronsted acid-base theory is the backbone of chemistry. This theory focuses mainly on acids and bases acting as proton donors or proton acceptors.

A^+ + B^- \rightarrow A-B

where A^+ is the Lewis Acid and B^- is the Lewis Base and A-B is the Covalent Bond.

Reaction of dissociation of HNO2 in H_2O is given as:

HNO_2 + H_2O \rightarrow H_3O^+ + NO_2^-

In this reaction O-atom has lone pair in water and therefore it accepts the proton from HNO_2 forming a Lewis Base.

6 0
3 years ago
Will give brainliest, like, 5 stars and 5 points to best answer
Oliga [24]

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6 0
3 years ago
A water bath in a physical chemistry lab is 1.75 m long, 0.730 m wide, and 0.650 m deep. If it is filled to within 2.27 inches f
german
It is 0.720 meters cause if the manufacturers of liters contain 2.27 inches it would make a deeply filled of 0.660
7 0
3 years ago
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According to the following reaction, how many grams of water are produced in the complete reaction of grams of sulfuric acid?
Natasha2012 [34]

This question requires the mass of sulfuric acid and a balance equation. The complete question is given below

Question:

According to the following reaction, how many grams of water are produced in the complete reaction of 24.1 grams of sulfuric acid?

H₂SO₄(aq) + Zn(OH)₂(s) ---------> ZnSO₄(aq) + 2H₂O(l)

Answer:

8.85 grams of water is obtained from 24.1 grams of sulfuric acid according to the given reaction

Explanation:

In this problem, the mass of water can be determined by using the balanced chemical equation.

Step 1: Write all data

Molar mass of water = 18 g

Molar mass of sulfuric acid = 98 g

Given mass of sulfuric acid = 24.1 g

Mass of water from reaction = ?

Step 2: Write statement for conversion

Given equation shows that

1 mole of H₂SO₄ gives 2 moles of water

Step 3: Convert moles into molar mass

Convert the moles into molar mass so, the statement becomes,

98 g of H₂SO₄ gives (2)(18) g of water

1 g of H₂SO₄ gives (2)(18)/98 g of water

Step 4: Use given data

24.1 g of H₂SO₄ gives (2)(18)(24.1)/98 g of water

amount of water = 8.85 g

5 0
3 years ago
How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
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