Answer:
2.813%
Explanation:
First we must take the average of the five mass measurements
Average value= sum of all five trials/ 5= 26.6 + 26.3 + 26.2 + 26.4 +26.1 /3 = 26.32mg
Now, we have been provided the accurate mass of the compound in the question as 25.6mg
Remember that relative percent error = Average value of measurement - True value of measurement / true value×100
Hence we have;
Percent error=
26.32 - 25.6/25.6 ×100
Percent error= 2.813%
The percentage error is 2.813%
The calculation is as follows:The Average value is
[tex]= (26.6 + 26.3 + 26.2 + 26.4 +26.1)\div 3[/tex]
= 26.32mg
Now the percent error should be
[tex]= 26.32 - 25.6\div 25.6 \times 100[/tex]
= 2.813%
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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . What is the theoretical yield of sodium sulfate formed from the reaction of of sulfuric acid and of sodium hydroxide
Answer:
71g (assuming experimental data)
Explanation:
The balanced equation for this reaction:
[tex]H_{2} SO_{4}(aq)[/tex] + [tex]2NaOH (s)[/tex] → [tex]Na_{2}SO_{4} (l)[/tex] + [tex]2H_{2} O (l)[/tex]
Molar mass of H2SO4 = 98.1 g/mol
molar mass of NaOH = 40g/mol
Molar mass of Na2SO4 = 142.04g/mol
⇒ 1 mole or 98.1g of H2SO4 will yield 1 ole of NaSO4; alternately, 2 moles or 49 ×2 = 80g of NaOH produces 1 mole of NaSO4.
Therefore, limiting reactant is NaOH.
Assuming actual experiment is 20g of NaOH,
1 mole - 40g
x moles - 20g = [tex]\frac{20}{40}[/tex] = 0.5 moles
⇒1 mole of Na2SO4 - 142.04g
∴ 0.5 moles = 142.04 × 0.5
= 71.02g
Answer:
71 grams
Explanation:
Match each ion with the noble guess whose electron configuration it shares
Answer:
These should be matched to either: Helium, Neon, or Argon. Na+. Be2+. Ca2+. Cl-.
Explanation:
According to the electronic configuration, Li[tex]^+[/tex] and Be[tex]^2+[/tex] have electronic configuration similar to helium while N[tex]^3-[/tex],Na[tex]^2+[/tex],O[tex]^2-[/tex] have electronic configuration similar to neon and rest have electronic configuration similar to argon.
What is electronic configuration?
Electronic configuration is defined as the distribution of electrons which are present in an atom or molecule in atomic or molecular orbitals.It describes how each electron moves independently in an orbital.
Knowledge of electronic configuration is necessary for understanding the structure of periodic table.It helps in understanding the chemical properties of elements.
Elements undergo chemical reactions in order to achieve stability. Main group elements obey the octet rule in their electronic configuration while the transition elements follow the 18 electron rule. Noble elements have valence shell complete in ground state and hence are said to be stable.
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The lava from a volcano can travel up to 100 km/hr if a village is located 3.6 mi from the volcano how many minutes will it take the lava to reach the village
Answer:
3.47 minutes
Explanation:
First we must convert the distance travelled from miles to kilometres
Since ;
1 mile= 1.6093 km
3.6 miles= 3.6 × 1.6093 km = 5.79 Km
Substitute this into
Speed= distance / time
Speed = 100Kmhr-1
Distance= 5.79 Km
Time= the unknown
Time = distance/speed
Time = 5.79/100
Time= 0.0579 hours
Since;
60 minutes make one hour
x minutes make 0.0579 hours
x = 60 × 0.0579 hours
x= 3.47 minutes
When 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1-butene is produced. When potassium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base
Answer:
The percentage mixture of 2-methyl-1-butene would be in between the 45% and 70%.
Explanation:
Potassium prop oxide is the intermediate base as compared to the potassium hydroxide which is less bulky strong base and potassium tert-butoxide which is bulky base. Bulky base can minimize the substitution reaction by causing hinders the approach of carbon attack and KOH is the strong base which less effective in minimizing the substitution reaction.
