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1 year ago

What is the molarity of 985 mL solution with 0.81 moles KF?

Chemistry

1 answer:

Mars2501 [29]1 year ago

8 0

Answer:

0.82 M

Explanation:

You have been given moles and volume (in mL). To find the molarity of the solution, you need to (1) convert mL to L and then (2) plug moles and volume into the molarity formula.

985 mL / 1,000 = 0.985 L

Molarity (M) = moles / volume (L)

Molarity = 0.81 moles / 0.985 L

Molarity = 0.82 M

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PH=3 المحلول حمضي او قاعده

Dominik [7]

Answer:

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Explanation:

pH=3 المحلول حمضي او قاعده

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2 years ago

1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g

Vitek1552 [10]

1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

so no. of moles of Hg(ClO4)2/ no. of moles of Na2S= 1/1 according to the eqn

so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

no. of moles of HgS produced= 0.10979mols

molar mass of HgS= 232.7g/mol 1d.p.

grams of solid produced= 232.7×0.10979= 25.5g (3s.f.)

2. reactant in excess is Hg(ClO4)2,

no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

3 0

3 years ago

Balance the following reactions using either the oxidation number method or the half-reaction method. Identify which element has

Marina86 [1]

$6Ce^{4+} + I^- + 6OH^-$ → $6Ce^{3+} + IO_3^- + 3H_2O$ is the balanced chemical equation.

<h3>What is a balanced chemical equation?</h3>

A balanced chemical reaction is an equation that has equal numbers of each type of atom on both sides of the arrow.

Half-reaction method:

Unbalanced chemical equation:

$Ce^{4+} + I^-$ → $Ce^{3+} + IO^{3-}$

Oxidation half-reaction:

$I^-+ 6OH^- - 6e-$ → I $O^{3-} + 3H_2O$

Reduction half-reaction:

$Ce4^+ + e^-$ → $Ce^{3+}$

Balanced chemical equation:

$6Ce^{4+} + I^- + 6OH^-$ → $6Ce^{3+} + IO^{3-} + 3H_2O$

Oxidation number method:

Unbalanced chemical equation:

$Ce^{4+} + I^-$ → $Ce^{3+} + IO^{3-}$

$I^{-1} -6e^-$ → $I^{+5}$

$Ce^{4+} + e^-$ → $Ce^{3+}$

Balanced chemical equation:

$6Ce^{4+} + I^{-1}$ → $6Ce^{3+} + I^{+5}$

$6Ce^{4+} + I^- + 6OH^-$ → $6Ce^{3+} + IO_3^- + 3H_2O$

Hence, $6Ce^{4+} + I^- + 6OH^-$ → $6Ce^{3+} + IO_3^- + 3H_2O$ is the balanced chemical equation.

Learn more about the balanced chemical equation here:

https://brainly.in/question/46754758

#SPJ1

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2 years ago

What is the correct formula for Elementium Nitride?

sashaice [31]

Answer:

AlN

Explanation:

have a great day

7 0

2 years ago

A quantity of 8.10 × 102 mL of 0.600 M HNO3 is mixed with 8.10 × 102 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of

victus00 [196]

Answer:

22.48°C is the final temperature of the solution.

Explanation:

Heat of neutralization of reaction , when 1 mol of nitric acid reacts= ΔH= -56.2 kJ/mol= -56200 J/mol

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of nitric acid = n

Volume of nitric acid solution = $8.10\times 10^2 mL= 0.81L$

Molarity of the nitric acid = 0.600 M

$n=0.600 M\times 0.81 L=0.486 mol$

Moles of barium hydroxide = n'

Volume of barium hydroxide solution = $8.10\times 10^2 mL= 0.81L$

Molarity of the barium hydroxide= 0.300 M

$n'=0.300 M\times 0.81 L=0.243 mol$

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_3+2H_2O$

According to reaction, 2 mol of nitric acid reacts with 1 mol of barium hydroxide .Then 0.486 mol of nitric acid will react with :

$\frac{1}{2}\times 0.486 mol=0.243 mol$ barium hydroxide.

Heat release when 0.486 mol of nitric acid reacted = Q

= ΔH × 0.486 = -56200 J/mol × 0.486 mol=-27,313.2 J

Heat absorbed by the mixture after reaction = Q' = -Q = 27,313.2 J

Volume of the nitric solution = 0.81 L = 810 mL

Volume of the ferric nitrate solution = 0.81 L = 810 mL

Total volume of the solution = 810 mL + 810 mL = 1620 mL

Mass of the final solution = m

Density of water = density of the final solution = d = 1 g/mL

$Mass=density\times Volume$

$m=1 g/ml\times 1620 ml=1620 g$

Initial temperature of the both solution were same = $T_1=18.46^oC$

Final temperature of the both solution will also be same after mixing= $T_2$

Heat capacity of the mixture = c = 4.184 J/g°C

Change in temperature of the mixture = ΔT = $(T_2-T_1)$

$Q=mc\Delta T=mc(T_2-T_1)$

$27,313.2 J= 1620 g\times 4.184 J/g^oC\times (T_2-18.46^oC)$

$T_2=22.49^oC$

22.48°C is the final temperature of the solution.

8 0

2 years ago