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Reil
2 months ago
12

Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to

neutralize 5.04×103 kg of sulfuric acid solution?
Chemistry
1 answer:
Anarel [2.9K]2 months ago
5 0

5.451 X 10³ kg of sodium carbonate is required to neutralize 5.04×10³ kg of sulfuric acid solution.

Explanation:

  • Sodium carbonate neutralizes sulfuric acid (H₂SO₄). This compound is derived from a strong base (NaOH) and a weak acid (H₂CO₃). The chemical equation for this neutralization process is represented as:

Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

  • The balanced equation indicates that one mole of Na₂CO₃ is needed to neutralize one mole of H₂SO₄.
  • Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol, while Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.
  • To neutralize 0.098 kg of H₂SO₄, the required Na₂CO₃ is 0.106 kg, thus, to neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ needed is 5.451 X 10³ kg.

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(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o
Tems11 [2777]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100

Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100

Percent Ionic character = 1 - e^(^-^0^.^1^) * 100

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0
3 months ago
"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92
lorasvet [2795]

Answer: The percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

From the information provided:

Let the fractional abundance for _{14}^{28}\textrm{Si} isotope be 'x'

  • For _{14}^{28}\textrm{Si} isotope:

Mass of _{14}^{28}\textrm{Si} isotope = 27.9769 amu

Percentage abundance of _{14}^{28}\textrm{Si} isotope = 92.22 %

Fractional abundance for _{14}^{28}\textrm{Si} isotope = 0.9222

  • For _{14}^{29}\textrm{Si} isotope:

Mass of _{14}^{28}\textrm{Si} isotope = 28.9764 amu

Percentage abundance for _{14}^{28}\textrm{Si} isotope = 4.68%

Fractional abundance of _{14}^{28}\textrm{Si} isotope = 0.0468

  • For _{14}^{30}\textrm{Si} isotope:

Mass of _{14}^{30}\textrm{Si} isotope = 29.9737 amu

Fractional abundance for _{14}^{30}\textrm{Si} isotope = x

  • The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309

To convert this fractional abundance into a percentage, multiply by 100:

\Rightarrow 0.0309\times 100=3.09\%

This shows that the percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

6 0
2 months ago
The nitrogen atom of NH2 would have The nitrogen atom of {\rm NH_2} would have blank electrons around the central nitrogen atom.
Alekssandra [3086]

Answer:

(a) The nitrogen atom in contains NH_2^-8 electrons surrounding the central nitrogen atom.

(b) The nitrogen atom in has NH_4^+8 electrons around the central nitrogen atom.

(c) The nitrogen atom in has NH_38 electrons surrounding the central nitrogen atom.

Explanation:

The Lewis dot structure illustrates the connections between atoms within a molecule, additionally showing unpaired electrons present in the molecule.In this structure, valence electrons are represented as 'dots'.

Now we will determine the number of electrons linked to the central nitrogen atom in the specified molecule.

(a) The identified molecule is,

NH_2^-Recognizing that nitrogen has '5' valence electrons while hydrogen has '1', we ascertain the total valence electrons in

= 5 + 2(1) + 1 = 8

According to the Lewis dot structure, this reveals 4 bonding and 4 non-bonding electrons.

NH_2^-From the Lewis structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

(b) The identified molecule is, NH_2^-

Knowing that nitrogen has '5' valence electrons and hydrogen has '1' valence electron, the total number of valence electrons for NH_4^+ = 5 + 4(1) - 1 = 8

In the Lewis dot structure, there are 8 bonding electrons and 0 non-bonding electrons.

This leads us to conclude that the nitrogen atom in

contains 8 electrons surrounding it.NH_4^+

(c) The identified molecule is,

NH_4^+

Recognizing nitrogen’s '5' valence electrons and hydrogen’s '1' brings the total valence electrons for

= 5 + 3(1) = 8NH_3Utilizing the Lewis dot structure indicates there are 6 bonding and 2 non-bonding electrons.

According to the Lewis dot structure, we conclude that the nitrogen atom in

has 8 electrons surrounding it.

NH_2^-

3 0
1 month ago
What is the correct name for the covalent compound N3Se5?
castortr0y [3046]

Result:

I believe it’s called Trinitrogen Pentaseleniumide

Explanation:

Tri means three

Penta means five

The second element concludes with -ide

6 0
2 months ago
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