First of all, you should know the valence electron of atom.
Then add charge, In case of -ve ion & subtract, in case of +ve ion.
A. True
Because Each element is made up of unique type of atom that has very specific properties
The percent yield of CO₂ is 93.3%.
<h3>What is the percent yield of CO₂?</h3>
The percent yield of a substance is given as follows:
- Percent yield = actual yield/theoretical yield * 100 %
The equation of the reaction is used to determine the theoretical yield.
- NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
Moe ratio of sodium bicarbonate and CO₂ is 1 : 1.
Given that sodium bicarbonate is the limiting reactant, the theoretical yield of CO₂ will be:
Moles of NaHCO₃ reacting = 2.01/84 = 0.0239 moles
Theoretical yield of CO₂ = 0.0239 moles * 22.4L/mol = 0.536 L
Actual yield = 0.50 L
Percent yield = 0.50/0.536 * 100%
Percent yield = 93.3%
In conclusion, the percent yield is the ratio of the actual yield and theoretical yield.
<em>Note that the complete question is given below:</em>
<em>Calculate your % yield of co2 in the reaction based on the grams of nahco3 being the limiting reagent in the reaction between 2.01 g of sodium bicarbonate and 24.6 mL of 1.5 M acetic acid? They produce 0.50 L of at s.t.p.</em>
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Answer:
1-Ethyl-3-methylidenecyclopentane
Step-by-step explanation:
Formula = C₈H₁₄. An alkane has formula C₈H₁₈. ∴ X contains 2 double bonds, 2 rings, or 1 ring and 1 double bond.
X absorbs only 1 mol of hydrogen. ∴ X contains 1 ring and 1 double bond.
Hydrogenation gives 1-ethyl-3-methylcyclopentane.
Ozonolysis gives formaldehyde, so X must contain a =CH₂ group.
Hydrogenation of X converted the =CH₂ to -CH₃.
X is 1-ethyl-3-methylidenecyclopentane.
You can see the reactions in the image below.
Answer:
Average atomic mass = 58.51047 amu
The symbol is 
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
For first isotope:
% = 68.274 %
Mass = 57.9353 amu
For second isotope:
% = 26.095 %
Mass = 58.9332 amu
For third isotope:
% = 1.134 %
Mass = 61.9283 amu
For fourth isotope:
% = 3.593 %
Mass = 63.9280 amu
For fifth isotope:
% = 0.904 %
Mass = 63.9280 amu
Thus,

Average atomic mass = 58.51047 amu
The symbol is 