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1 year ago

Two 25.0-mL aqueous solutions, labeled A and B, contain the ions indicated:

Chemistry

1 answer:

asambeis [7]1 year ago

4 0

The additional volume of HCl which must be added to reach to the equivalence point is 8.33 mL

The moles of HCl which is required to reach the equivalence point can be calculated in the way as follows.

Moles of HCl can be calculated as

Moles of HCl = 0.004 moles of Ca (OH) 2 × 2 moles of HCl / 1 moles of Ca (OH) 2

= 0.008 moles of HCl

The volume of HCl which is required to reach the equivalence point can be calculated in the way given as follows.

Volume of HCl required= 0.008 moles of HCl × 1 L / 0.24 moles of HCl × 1 ml / 10 -³ L

= 33.33 ml

The additional volume of HCl calculated as

Additional volume = required volume – actual volume

= 33.33 mL – 25 mL

= 8 . 33 mL

Thus, we calculated that the additional volume of HCl which must be added to reach to the equivalence point is 8.33 mL.

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Calculate the percentage of water in the hydrate manganese (ii) nitrate tetrahydrate

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Answer:

Divide the mass of the water lost by the mass of hydrate and multiply by 100.

Explanation:

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5Br−+BrO3−+6H+→3Br2+3H2O

sashaice [31]

Explanation :

The balanced chemical reaction is,

$5Br^-+BrO_3^-+6H^+\rightarrow 3Br_2+3H_2O$

The expression for the rates of consumption of the reactants are:

The rate of consumption of $Br^-$ = $-\frac{1}{5}\frac{d[Br^-]}{dt}$

The rate of consumption of $BrO_3^-$ = $-\frac{d[BrO_3^-]}{dt}$

The rate of consumption of $H^+$ = $\frac{1}{6}\frac{d[H^+]}{dt}$

The expression for the rates of formation of the products are:

The rate of consumption of $Br_2$ = $+\frac{1}{3}\frac{d[Br_2]}{dt}$

The rate of consumption of $H_2O$ = $+\frac{1}{3}\frac{d[H_2O]}{dt}$

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3 years ago

If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?

Hitman42 [59]

Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.

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3 years ago

Atomic number and mass of O

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Atomic number is 8 and atomic mass is taken as 16 amu

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3 years ago

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Part A of the lab involved adding 4 mL increments of distilled water to 5.00 mL of antimony trichloride solution. The antimony t

Mrrafil [7]

Answer:

0.0238M SbCl3, 1.07M H+, 1.14M Cl-

Explanation:

The total volume of the solution is:

4mL + 5.00mL + 12.0mL = 21mL

As the volume of the SbCl3 is 5.00mL, the dilution factor is:

21mL / 5.00mL = 4.2 times

The concentration of SbCl3 is:

0.10M SbCl3 / 4.2 times = 0.0238M SbCl3

The concentration of H+ = [HCl]:

4.5M / 4.2 times = 1.07M H+

The initial concentration of Cl- is:

3 times SbCl3 + HCl = 0.10M*3 + 4.5M =

<em>3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-</em>

4.8M Cl- / 4.2 times = 1.14M Cl-

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2 years ago