Answer:
BCI3 is a non polar compound because there is no neutral in it
Answer: B. It’s a dilute strong base.
Explanation:
1) Definition of acids and bases: as per Bronsted-Lowry model, an acid is a substance that donates hydrogen ions and a base is a substance that accepts hydrogen ions.
Ca(OH)₂ does not have hydrogen ions to donate, but it can accept hydrogen ions to form H₂O according to this equation: H⁺ + OH⁻ → H₂O.
Hence, Ca(OH)₂ is a base.
2) Definition of strong base: a strong base is a base that dissociates completely into metallic and hydroxide ions in aqueous solutions, while a weak base dissociates partially.
Hence, Ca(OH)₂ is a strong base.
3) Definition of dilute: it refers to a solution meaning that the substance is not pure and the concentration is low. Since, the solution the Ca(OH)₂ is 0.02 M means that it is dilute.
Therefore, we have found that the description of 0.02 M Ca(OH)₂ is that is is a dilute strong base (option B).
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i think thats right
Answer:
As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Explanation:
Absorbance of light by a reagent of concentration c, is given as
A = εcl
A = Absorbance
ε = molar absorptivity
c = concentration of reagent.
l = length of light path or length of the solution the light passes through.
So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.
But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.
Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.
Hope this Helps!!!
In one mole of talc, we observe that there are:
3 moles of Mg
4 moles of Si
2 moles of H
12 moles of O
The molar ratio of O to Mg is then:
12 moles of O : 3 moles of Mg = 4 : 1
Therefore, if 6.1 moles of Mg are present, the moles of O are:
4 * 6.1 = 24.4 moles of O
