Answer:
a)
⇒
⇒
b)
⇒
⇒
Explanation:
A)
Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:
⇒
The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:
⇒
b)
Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:
⇒
Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:
⇒
Answer:
Left hand side:-
Carbon - 12
HYdrogen - 28
Oxygen - 38
Right hand side:-
Carbon - 12
Hydrogen - 28
Oxygen - 38
Since, the number of atoms each side are equal, the reaction is balanced.
Explanation:
The given reaction is:-

Left hand side:-
Carbon - 12
HYdrogen - 28
Oxygen - 38
Right hand side:-
Carbon - 12
Hydrogen - 28
Oxygen - 38
<u>Since, the number of atoms each side are equal, the reaction is balanced.</u>
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
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The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
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