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Ratling [72]
3 years ago
5

What volume (in mLmL) of a 0.175 MHNO3MHNO3 solution is required to completely react with 44.2 mLmL of a 0.108 MNa2CO3MNa2CO3 so

lution according to the following balanced chemical equation? Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Answer:

54.5 mL

Explanation:

Let's consider the following balanced equation.

Na₂CO₃(aq) + 2 HNO₃(aq) → 2 NaNO₃(aq) + CO₂(g) + H₂O(l)

44.2 mL of 0.108 M Na₂CO₃ solution react. The moles of Na₂CO₃ that react are:

0.0442 L × 0.108 mol/L = 4.77 × 10⁻³ mol

The molar ratio of Na₂CO₃ to HNO₃ is 1:2. The moles of HNO₃ that reacted are  2/1 × 4.77 × 10⁻³ mol = 9.54 × 10⁻³ mol

9.54 × 10⁻³ moles of HNO₃ are in a 0.175 M solution. The required volume of HNO₃ is:

9.54 × 10⁻³ mol × (1 L/0.175 mol) × (1000 mL/1 L) = 54.5 mL

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43 milliliters of water weighs 43 g. what is the density of the water?
Anastaziya [24]

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Hello,

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4 0
3 years ago
Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

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4 years ago
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