The product and balanced net ionic equations for the following reactions are SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂.
Ionic equations are those equations that happened in an aqueous solution. The chemical equation is expressed in an electrolyte solution is expressed and dissociates ions.
In these reactions, each element or ion is dissociated into differently charged ions in a solution. Each of the ions is shown with different charges.
SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂. In the reaction, the tin chloride, and potassium magnesium oxide. It dissociates into charged ions, like potassium chloride and tin magnesium oxide. The chlorine will acquire a negative charge and magnesium oxide get a positive charge.
Thus, the net ionic equation is SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂.
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Answer:
The answer to your question is:
Explanation:
Reaction
SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂
1 ---- Sn ---- 1
2 ---- K ----- 2
2 ---- Mn ---- 2
8 ---- O ---- 8
2 ---- Cl ---- 2
The balanced chemical equation for the reaction between hydrogen sulfide and oxygen is:
2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g)
We can interpret this to mean:
3moles of oxygen and_______moles of hydrogen sulfide react to produce______moles of water and_______ moles of sulfur dioxide.
Answer:
1. The coefficients are: 1, 3, 2
2. From the balanced equation, we obtained the following:
3 moles oxygen, O2 reacted.
2 moles of Hydrogen sulfide, H2S reacted.
2 moles of water were produced.
2 moles of sulphur dioxide, SO2 were produced.
Explanation:
1. Determination of the coefficients of the equation.
This is illustrated below:
P2O5(s) + H2O(l) <==> H3PO4(aq)
There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:
P2O5(s) + H2O(l) <==> 2H3PO4(aq)
There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:
P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)
Now the equation is balanced.
The coefficients are: 1, 3, 2.
2. We'll begin by writing the balanced equation for the reaction. This is given below:
2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)
From the balanced equation above,
3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.
In the balanced chemical equation provided, 3 moles of oxygen react with 2 moles of hydrogen sulfide to produce 2 moles of water and 2 moles of sulfur dioxide.
When the balanced chemical equation 2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g) is considered, we can deduce that 3 moles of oxygen and 2 moles of hydrogen sulfide react together in this reaction. The products of this chemical reaction are 2 moles of water and 2 moles of sulfur dioxide. Each of these quantities is directly inferred from the coefficients in front of each compound in the balanced chemical equation.
Answer:
1.7 × 10⁴ J
Explanation:
Step 1: Calculate the heat required to raise the temperature of ice from -15 °C to 0°C
We will use the following expression.
Q₁ = c(ice) × m × ΔT
Q₁ = 2.03 J/g.°C × 25 g × [0°C - (-15°C)] = 7.6 × 10² J
Step 2: Calculate the heat required to melt 25 g of ice
We will use the following expression.
Q₂ = C(fusion) × m
Q₂ = 80. cal/g × 25 g × 4.184 J/1 cal = 8.4 × 10³ J
Step 3: Calculate the heat required to raise the temperature of water from 0°C to 75 °C
We will use the following expression.
Q₃ = c(water) × m × ΔT
Q₃ = 4.184 J/g.°C × 25 g × (75°C - 0°C) = 7.8 × 10³ J
Step 4: Calculate the total heat required
Q = Q₁ + Q₂ + Q₃
Q = 7.6 × 10² J + 8.4 × 10³ J + 7.8 × 10³ J = 1.7 × 10⁴ J
Answer:
1.62 g
Explanation:
Given that:
Concentration of HCN = 0.119 M
Assuming the ka 4.00 × 10⁻¹⁰
The pKa of HCN (hydrocyanic acid) = -log (Ka)
= - log ( 4.00 × 10⁻¹⁰)
= 9.398
pH of buffer = 8.809
Using Henderson Hasselbach equation:
[CN^-] = 0.2576[HCN]
[CN^-] = 0.2756 (0.119) L
[CN^-] = 0.033 M
∴
The amount of NaCN (sodium cyanide) is calculated as follows:
= 1.62 g
Answer:
The free energy = -20.46 KJ
Explanation:
given Data:
Pb²⁺ = 0.750 M
Br⁻ = 0.232 M
R = 8.314 Jk⁻¹mol⁻¹
T = 298K
The Gibb's free energy is calculated using the formula;
ΔG = ΔG° + RTlnQ -------------------------1
Where;
ΔG° = standard Gibb's freeenergy
R = Gas constant
Q = reaction quotient
T = temperature
The chemical reaction is given as;
Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)
The ΔG°f are given as:
ΔG°f (PbBr₂) = -260.75 kj.mol⁻¹
ΔG°f (Pb²⁺) = -24.4 kj.mol⁻¹
ΔG°f (2Br⁻) = -103.97 kj.mol⁻¹
Calculating the standard gibb's free energy using the formula;
ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)
Substituting, we have;
ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]
ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)
= -260.75 + 232.34
= -28.41 kj
Calculating the reaction quotient Q using the formula;
Q = 1/[Pb²⁺ *(Br⁻)²]
= 1/(0.750 * 0.232²)
= 24.77
Substituting all the calculated values into equation 1, we have
ΔG = ΔG° + RTlnQ
ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)
= -28.41 +7.95
= -20. 46 kJ
Therefore, the free energy of reaction = -20.46 kJ
To calculate the reaction free energy ΔG for this reaction, we need to use the standard free energy of formation values given in a data tab, the stoichiometry of the reaction, and the specific conditions of the reaction, including the concentrations of Pb2+ and Br−. After a series of calculations, we will get the ΔG value in joules, which can be converted to kilojoules.
The task here is to calculate the reaction free energy ΔG for the Pb2+(aq) + 2Br−(aq) = PbBr2(s) reaction at 25.0°C. From the given information, we can start by calculating the number of moles of PbBr2 from its mass. Then, referring to the thermodynamic data tab of the ALEKS, we find the standard free energy of formation (ΔGf°) values for Pb2+(aq), Br−(aq), and PbBr2(s). Now, we can use these values and the definition of ΔG for a reaction in terms of ΔGf° values and stoichiometry.
ΔG = ΣΔGf°(products) - ΣΔGf°(reactants).
Note that the equation must be balanced so each ΔGf° value is multiplied by the stoichiometric coefficient of that substance in the reaction. It is also important to remember to convert the answer to kilojoules if the ΔGf° values are given in joules/mole. Lastly, the concentrations of Pb2+ and Br− are included in the reaction quotient Q to show the reaction's non-standard conditions.
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emperature. What is the independent variable in his experiment?
a. The unknown substance, because it's the only thing he changed
b.The temperature, because it's the only thing he changed
Answer:
a
Explanation: