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11111nata11111 [884]
3 years ago
6

If you were to add a proton to a neutral atom what would change

Chemistry
1 answer:
viktelen [127]3 years ago
8 0

It would change the charge, a neutral atom has zero charge but a proton has a positive charge.  So 0 charge + 1 positive charge = 1 positive charge.

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How does Barium obey the octet rule when reacting
Kryger [21]

Answer:

it gives up electrons B. because the bond between the electrons is not very strong due to how few of them there are

Explanation:

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PLEASE SOMEONE! The density of the acetic acid solution is 1.05 g/mL. Calculate the %(m/m) of the acetic acid solution (convert
tigry1 [53]

Answer:

The Percentage concentration of acetic acid = 1.39 %

Explanation:

Density of acetic acid solution = 1.05 g/mL

Volume of acetic acid solution = 0.1 L = 100 mL

From the formula, Density = mass / volume; mass = density × volume

Mass of acetic acid solution = 1.05 g/mL × 100 mL = 105 g

Molar concentration of acetic acid solution = 0.243 mol/L

Molar mass of acetic acid, CH₃COOH = (12 × 2 + 1 ×4 + 16 ×2) = 60 g/mol

From the formula, mass concentration = molar concentration × molar mass

Mass concentration of acetic acid, CH₃COOH = 0.243 mol/L × 60 g/mol = 14.58 g/L

In one liter of acetic acid solution, there are 14.58 g of acetic acid. Therefore, in 0.1 L, there will be 14.58 × 0.1 = 1.458 g of acetic acid.

Percentage concentration of acetic acid = mass of acetic acid / mass of acetic acid solution × 100%

Percentage concentration of acetic acid = (1.458 / 105) × 100% = 1.39 %

The Percentage concentration of acetic acid = 1.39 %

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3 years ago
What are the three conditions that can cause a chemicl reactions, give examples?<br> Help me plz
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A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would
Olenka [21]

Answer:

C. Ca_3(PO_4)_2  will precipitate out first

the percentage of Ca^{2+}remaining =  12.86%

Explanation:

Given that:

A solution contains:

[Ca^{2+}] = 0.0440 \ M

[Ag^+] = 0.0940 \ M

From the list of options , Let find the dissociation of Ag_3PO_4

Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}

where;

Solubility product constant Ksp of Ag_3PO_4 is 8.89 \times 10^{-17}

Thus;

Ksp = [Ag^+]^3[PO_4^{3-}]

replacing the known values in order to determine the unknown ; we have :

8.89 \times 10 ^{-17}  = (0.0940)^3[PO_4^{3-}]

\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}  = [PO_4^{3-}]

[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}

[PO_4^{3-}] =1.07 \times 10^{-13}

The dissociation  of Ca_3(PO_4)_2

The solubility product constant of Ca_3(PO_4)_2  is 2.07 \times 10^{-32}

The dissociation of Ca_3(PO_4)_2   is :

Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}

Thus;

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33} = (0.0440)^3  [PO_4^{3-}]^2

\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}=   [PO_4^{3-}]^2

[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}

[PO_4^{3-}]^2 = 2.43 \times 10^{-29}

[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}

[PO_4^{3-}] =4.93 \times 10^{-15}

Thus; the phosphate anion needed for precipitation is smaller i.e 4.93 \times 10^{-15} in Ca_3(PO_4)_2 than  in  Ag_3PO_4  1.07 \times 10^{-13}

Therefore:

Ca_3(PO_4)_2  will precipitate out first

To determine the concentration of [Ca^+] when  the second cation starts to precipitate ; we have :

Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2

2.07 \times 10^{-33}  = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2

[Ca^{2+}]^3 =  \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}

[Ca^{2+}]^3 =1.808 \times 10^{-7}

[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}

[Ca^{2+}] =0.00566

This implies that when the second  cation starts to precipitate ; the  concentration of [Ca^{2+}] in the solution is  0.00566

Therefore;

the percentage of Ca^{2+}  remaining = concentration remaining/initial concentration × 100%

the percentage of Ca^{2+} remaining = 0.00566/0.0440  × 100%

the percentage of Ca^{2+} remaining = 0.1286 × 100%

the percentage of Ca^{2+}remaining =  12.86%

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True... there are better ways using less materials and printing is that
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