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.
Answer:
Explanation:
Remark
This is one of those questions that you need the choices. You can't tell what you should enter. For example, in scientific notation, you would get 1.5*10^7.
Or you could keep it as an integer and round to two places 150,000,000,
I would pick scientific notation if you know how to use it. Otherwise use the interger format.
Tetrahedral arrangement is resulted upon mixing one s and three p atomic orbitals, resulting in 4 hybridized
orbitals →
hybridization.
<h3>What is
orbital hybridization?</h3>
In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.
For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.
Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.
Learn more about Hybridization
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To find - Identify what kind of ligand (weak or strong), what kind
of wavelength (long or short), what kind of spin (high spin or
low spin) and whether it is paramagnetic or diamagnetic for
the following complexes.
1. [Mn(CN)6]4-
2. [Fe(OH)(H2O)5]2
3. [CrCl4Br2]3-
Step - by - Step Explanation -
1.
[Mn(CN)⁶]⁴⁻ :
Ligand - Strong
Wavelength - Short
Spin - Low spin
Number of unpaired electrons = 1 ∴ paramagnetic.
2.
[Fe(OH)(H₂O)₅]²⁺ :
Ligand - Weak ( both OH⁻ and H₂O )
Wavelength - Long
Spin - High spin
Number of unpaired electrons = 5 ∴ paramagnetic.
3.
[CrCl₄Br₂]³⁻ :
Ligand - Weak ( both Br⁻ and Cl⁻ )
Wavelength - Long
Spin - High spin
Number of unpaired electrons = 3 ∴ paramagnetic.
Answer: with? i can help:))
Explanation:
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