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lana66690 [7]
3 years ago
7

- The subatomic particle possessing the largest

Chemistry
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:

B. electrons possess the largest charge-to-mass ratio among the subatomic particles listed in the four choices.

Explanation:

Consider the mass of each particle. Express the masses in atomic mass units:

  • Protons: approximately 1.007 amu each;
  • Neutrons: approximately 1.009 amu each;
  • Electrons: approximately 0.0005 amu each.

Similarly, consider the charge on each particle. Express the charges in multiples of the fundamental charge:

  • Protons: +1 e;
  • Neutrons: 0;
  • Electrons: -1 e.

Calculate the charge-to-mass ratio for the three species:

  • Protons: approximately \rm 0.99\; e\cdot amu^{-1};
  • Neutrons: 0;
  • Electrons: approximately \rm 1,800\;e \cdot amu^{-1}.

Almost all nuclei contain protons and neutrons. The only exception is the hydrogen-1 nucleus, which contains only one proton and no neutron. The mass of the nucleus is approximately the same as the sum of its components' masses. The extra neutron will only add to the mass of the nucleus (the denominator) without contributing to the charge (the numerator.) As a result, the charge-to-mass ratio of nuclei will be positive but no greater than the charge-to-mass ratio of protons.

Among the particles in the four choices, the charge-to-mass ratio is the greatest for electrons.

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G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu
Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄Ag^{+} + Br^{-1}.

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains \frac{0.6485}{45}×1000 = 14.411 gm-mole. Thus the solubility product K_{sp}of AgBr = [Ag^{+}]×Br^{-}.

Or, 5.0×10^{-13} = S^{2} (the given value of solubility product of AgBr is 5.0×10^{-13} and the charge of the both ions are same).

Thus S = (5.00×10^{-13})^{1/2} = 7.071×10^{-7} g/mL.

Thus the concentration of Br^{-1} or HBr is 7.071×10^{-7} g/mL.

4 0
3 years ago
What is the concentration of a solution with a volume of 2.5 liters containing 600 grams of calcium phosphate?​
trapecia [35]

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution  = 2.5L

Mass of Calcium phosphate  = 600g

Unknown:

Concentration  = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

        Formula of calcium phosphate  = Ca₃PO₄

         molar mass = 3(40) + 31 + 4(16) = 215g/mol  

Number of moles of  Ca₃PO₄  = \frac{600}{215}   = 2.79moles

   Now;

  Concentration  = \frac{Number of moles }{volume }  

 Concentration  = \frac{2.79}{2.5}   = 1.12M

7 0
3 years ago
Suppose of nickel(II) chloride is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of chl
stich3 [128]

Answer: Molarity of chloride anion = 0.32 M

<em>Note: the question is missing some values. The full question is given below;</em>

<em>Suppose 7.26 g of nickel(II) chloride is dissolved in 350 mL of a 0.50 M aqueous solution of potassium carbonate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the nickel(II) chloride is dissolved in it. Be sure your answer has the correct number of significant digits.</em>

Explanation:

Molarity or molar concentration is the number of moles (mol) of component per volume (liters) concentration of solution in mol/L or M

The mass of nickel (II) chloride is 7.26 g.

The volume of potassium carbonate is 350 mL = 0.35 L

The molarity of potassium carbonate solution is 0.50 M

The reaction of nickel (II) chloride and potassium carbonate is given below.

NiCl₂(aq) + KCO₃(aq) --------> KCl(aq) +NiCO₃(s)

The dissociation of nickel (II) chloride is given below.

NiCl₂   -----> Ni²⁺ + 2Cl⁻

The molar mass of nickel (II) chloride is  129.6 g/mol

The moles of nickel (II) chloride can be calculated by the formula given below;

No of moles  = mass(g) / molar mass (g/mol)

No of moles = 7.26 / 129.6 = 0.056 moles

Therefore, molarity of NiCl₂ = 0.056 moles/ 0.35 L = 0.16 M

The molarity of 1 mole nickel (ii) chloride is 0.16 m and according to dissociation of nickel (II) chloride, 1 mole of nickel (II) chloride gives 2 moles of chloride anion.

Therefore, the molarity of chloride anion = 0.16 * 2 = 0.32 M

3 0
3 years ago
HELP PLEASE
ololo11 [35]
Third choice because the others make no sense
8 0
2 years ago
Read 2 more answers
PLEASE HELP I'M SO DOOMED
BartSMP [9]

Answer:

D

Explanation:

31 / 2.8 = 11.0714286 L per mole of helium

3.5 / 4 = 0.875 moles

2.8 + 0.875 = 3.675 moles

11.0714286 x 3.675 = 40.6875 L

3 0
3 years ago
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