Answer:
1. 2Al + 3I2 —> Al2I6
2. 0.555mol of I2
Explanation:
1. Al + I2 —> Al2I6
Observing the above equation, there are 2 atoms of Al on the right side and 1 on the left side. To balance it, put 2 in front of Al as shown below:
2Al + I2 —>Al2I6
Also, there are 6 atoms of I on the right side and 2 on the left side. To balance it, put 3 in front I2 as shown below:
2Al + 3I2 —>Al2I6
2. Molar Mass of Al = 27g/mol
Mass of Al = 10g
n = Mass /Molar Mass
n = 10/27 = 0.37mol
From the equation,
2moles of Al reacted with 3 moles of I2.
Therefore, 0.37mol of Al will react with = (0.37 x 3)/2 = 0.555mol of I2
The balanced chemical equation is:
2 Al(s) + 3 I₂(s) → Al₂I₆(s)
0.557 moles of iodine react with 10.0 g of aluminum.
Let's consider the following unbalanced equation.
Al(s) + I₂(s) → Al₂I₆(s)
We will balance it using the trial and error method. We can get the balanced equation by multiplying Al by 2 (balance Al atoms) and I₂ by 3 (balance I atoms).
2 Al(s) + 3 I₂(s) → Al₂I₆(s)
The molar mass of Al is 26.98 g/mol. The moles corresponding to 10.0 g of Al are:
The molar ratio of Al to I₂ is 2:3. The moles of I₂ that react with 0.371 moles of Al are:
The balanced chemical equation is:
2 Al(s) + 3 I₂(s) → Al₂I₆(s)
0.557 moles of iodine react with 10.0 g of aluminum.
You can learn more about stoichiometry here: brainly.com/question/9743981
Answer:
Heterogeneous Mixture. Have a good day! =)
Explanation:
Answer:
Three atoms are attached to the central atom in NF3.
Explanation:
The central atom is always regarded as the atom having the least electronegativity in a molecule or ion. We can decide on what atom should be the central atom by comparing the relative electro negativities of the atoms in the molecule or ion.
If we consider NF3, we can easily see that nitrogen is less electronegative than fluorine, hence nitrogen is the central atom in the molecule. We can also observe from the molecular model that three atoms of fluorine were attached to the central atom. Hence there are three atoms attached to the central atom in the molecule NF3.
Answer:
The mixture is not in equilibrium, the reaction will shift to the left.
Explanation:
Based on the equilibrium:
Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺
kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]
Where [] are concentrations at equilibrium. The reaction is in equilibrium when the ratio of concentrations = kc
Q is the same expression than kc but with [] that are not in equilibrium
Replacing:
Q = [10.0M] [1.0M] / [0.1M] [0.1M]
Q = 1000
As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc
The mixture's equilibrium status can be determined by comparing the reaction quotient (Q) with the equilibrium constant (Kc). If Q < Kc, the reaction proceeds to the right (products) to achieve equilibrium. If Q > Kc, the reaction proceeds to the left (reactants) to achieve equilibrium.
To determine if the mixture is initially at equilibrium, we need to calculate and compare the reaction quotient (Q) and the equilibrium constant (Kc) of the reaction. The reaction quotient is a measure of the relative concentrations of products and reactants at any point in time, whereas Kc, is the measure of these concentrations only at equilibrium.
Assuming that the reaction in question is: Fe3+ + HSCN ↔ FeSCN2+ + H + . In this case,
Q = [FeSCN2+]/[Fe3+][HSCN] = 10 / (0.1 * 0.1) = 1000. If Kc is less than 1000, the reaction is not at equilibrium and will need to proceed to the left (reactants) to reach equilibrium. Conversely, if Kc is greater than 1000, the reaction is not at equilibrium and will need to proceed to the right (products).
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Explanation:
Ionic equation
NaCl(aq) --> Na+(aq) + Cl-(aq)
Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)
In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.
Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)
= 142 g/mol
Molecular weight of NaCl = 23 + 35.5
= 58.5 g/mol
Masses
% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100
= 46/142 * 100
= 32.4%
% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100
= 23/58.5 * 100
= 39.3%
Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.
You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.
No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.
For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.
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