How many grams of CO are needed to react with 3.34 g of Fe2O3?

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

2 years ago

Question 2

Alenkinab [10]

Answer:

1.53 atm

Explanation:

From the question given above, the following data were obtained:

Volume = constant

Initial pressure (P₁) = stp = 1 atm

Initial temperature (T₁) = 273 K

Final temperature (T₂) = 144 °C = 144 °C + 273 = 417 K

Final pressure (P₂) =?

Since the volume is constant, the final pressure can be obtained as follow:

P₁ / T₁ = P₂ / T₂

1 / 273 = P₂ / 417

Cross multiply

273 × P₂ = 417

Divide both side by 273

P₂ = 417 / 273

P₂ = 1.53 atm

Therefore, the final pressure (i.e the pressure inside the hot water bottle) is 1.53 atm.

8 0

2 years ago

Show the equation how many grams of Mg(OH)2 would be produced from 4 mol KOH?

Sindrei [870]

Mole ratio:

MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl

2 moles KOH ---------------- 1 mole Mg(OH)₂
4 moles KOH ------------------- moles Mg(OH)₂

moles Mg(OH)₂ = 4 x 1 / 2

= 2 moles of Mg(OH)₂

molar mass Mg(OH)₂ = 58g/mol

mass of Mg(OH)₂ = n x mm

mass of Mg(OH)₂ = 58 x 2

= 116 g of Mg(OH)₂

hope this helps!

8 0

2 years ago

If the reactants contain one chlorine molecule, do you know how many chlorine atoms will be in the product? Include conversion o

max2010maxim [7]

Answer:

The number of Chlorine atoms in the product is 2.

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed in a chemical reaction.

The reactants contain one chlorine molecule( $Cl_{2}$ ) which has two chlorine atoms.