Mass percentage of a solution is the amount of solute present in 100 g of the solution.
Given data:
Mass of solute H2SO4 = 571.3 g
Volume of the solution = 1 lit = 1000 ml
Density of solution = 1.329 g/cm3 = 1.329 g/ml
Calculations:
Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g
Therefore we have:
571.3 g of H2SO4 in 1329 g of the solution
Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987
Mass percentage of H2SO4 (%w/w) is 42.99 %
Answer:
Explanation:
Using the necessary reagents to faciliate the synthesis of the organic compounds as shown in the attached file.
Use the equation q=ncΔT.
q= heat absorbed our released (in this case 1004J)
n= number of moles of sample ( in this case 2.08 mol)
c=molar heat capacity
ΔT=change in temperature (in this case 20°C)
You have to rewrite the equation for c.
c=q/nΔT
c=1004J/(2.08mol x 20°C)
c=24.1 J/mol°C
I hope this helps
Answer:
The correct answer is thermophiles.
Explanation:
Thermus aquaticus are heat resistant bacteria because these bacteria can survive under adverse environmental conditions like high temperature.
These bacteria belong to one of the most heat-loving groups of extremophiles that are thermophiles. Thermophiles are present in volcanic soil, geysers and around deep-sea vents where the temperature is extremely high.
Thermus aquaticus bacteria is used to manufacture an enzyme called Taq DNA polymerase, which is heat resistant and also an important factor in molecular biology.
Mass of CO₂ evolved : 0.108 g
<h3>Further explanation</h3>
Given
1.205g sample, 36% MgCO3 and 44% K2CO3
Required
mass of CO2
Solution
0.36 x 1.205 g=0.4338 g
mass C in MgCO₃(MW MgCO₃=84 g/mol, Ar C = 12/gmol)
= (12/84) x 0.4338
= 0.062 g
0.44 x 1.205 g = 0.5302 g
Mass C in K₂CO₃(MW=138 g/mol) :
= (12/138) x 0.5302
= 0.046 g
Total mass Of CO₂ :
= 0.062 + 0.046
= 0.108 g
