345 NH3 in moles is: 20.2577490294015 moles.
Answers are:
1) The balanced oxidation half reaction: 2I⁻(aq) → I₂(s) + 2e⁻.
Iodine is oxidized (lost electrons) from -1 to neutral charge (0).
2) The balanced reduction half-reaction: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻.
Hydrogen is reduced (gain electrons) from +1 to neutral charge.
3) The oxidation <span>reaction takes place at the anode.</span>
A liquid has definite volume but no definite shape .
<span>The </span>abundance of a chemical element<span> is a measure of the </span>occurrence<span> of the </span>element<span> relative to all other elements in a given environment. Abundance is measured in one of three ways: by the </span>mass-fraction<span> (the same as weight fraction); by the </span>mole-fraction<span> (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the </span>volume-fraction<span>. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and </span>ideal gas<span> mixtures. Most abundance values in this article are given as mass-fractions.
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Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
