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- A)Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)
- 2KBr(aq)+BaI2(aq) → 2KI(aq)+BaBr2(s)
- B)Balance the Chemical Equation for the reaction of calcium carbonate with hydrochloric acid:
- CaCO3+ HCl -> CaCl2 + CO2 + H2O To balance chemical equations we need to look at each element individually on both sides of the equation. calcium carbonate is a chemical compound with the formula CaCO3.
<em><u>M</u></em><em><u>a</u></em><em><u>r</u></em><em><u>k</u></em><em><u> </u></em><em><u>m</u></em><em><u>e</u></em><em><u> </u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>b</u></em><em><u>r</u></em><em><u>a</u></em><em><u>i</u></em><em><u>n</u></em><em><u>l</u></em><em><u>i</u></em><em><u>s</u></em><em><u>t</u></em>
Resonance, leaving group, carbonyl carbon delta+, and steric effect is the most crucial variables that affect the relative reactivity of a functional group containing a carbonyl in an addition or substitution process.
Discussion:
1. Carbonyl Carbon Delta+: The carbonyl group becomes more electrophilic and accelerates nucleophilic assault when the carbonyl carbon delta+ is bigger.
2. Resonance: When the carbonyl is transformed into the tetrahedral adduct, it may be lost. Loss of resonance increases the energy of the transition state for this nucleophilic assault because resonance has the function of stabilizing. Therefore, a carbonyl functional group's resistance to nucleophilic attack increases as resonance in the group increases in importance.
3. Leaving group: Tetrahedral adduct fragmentation is encouraged by a better LG.
4. Steric effects: The nucleophilic attack on carbonyl carbon is delayed when sterically impeded.
Learn more about carbonyl here:
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Answer:
1) 1.15 mol
2) M=0.45
3) 22.5 mL
4) 6.25 mL
Explanation:
1)
550 mL= 0.55 L
M= mol solute/ L solution
mol solute= M * L solution
mol solute= (2.1 M * 0.55 L ) M=1.15 mol solute
2)
155 mL = 0.155 L
80 g -> 1 mol NH4NO3
5.61 g -> x
x= (5.61 g * 1 mol NH4NO3)/80 g x= 0.07 mol NH4NO3
M=(0.07 mol NH4NO3)/0.155 L M=0.45
3) M1V1=M2V2
V1= M2V2/M1
V1= (0.500 M * 0.225 L)/5.00 M V1=0.0225 L =22.5 mL
4) M1V1=M2V2
V1= M2V2/M1
V1= (0.25 M * 0.45 L)/ 18.0 M
V1=6.25 x 10^-3 L = 6.25 mL
Answer is: D. Cl (chlorine).
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Barium, potassium and arsenic are metals (easily lost valence electrons), chlorine is nonmetal (easily gain electrons).
Alkaline metals (in this example, potassium) have lowest ionizations energy and easy remove valence electrons (one electron), earth alkaline metals (in this example, barium) have higher ionization energy than alkaline metals, because they have two valence electrons.
Nonmetals (in this example chlorine) are far right in the main group and they have highest ionization energy, because they have many valence electrons.
Answer:
<h2>227.27 mL</h2>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question we have
We have the final answer as
<h3>227.27 mL</h3>
Hope this helps you
