Answer: 0.4533mol/L
Explanation:
Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol
68g of CaCO3 dissolves in 1.5L of solution.
Xg of CaCO3 will dissolve in 1L i.e
Xg of CaCO3 = 68/1.5 = 45.33g/L
Molarity = Mass conc.(g/L) / molar Mass
Molarity = 45.33/100 = 0.4533mol/L
3 answer for number 3.4 is the answer for number 4
<span>In this work the performance of gypsum plaster and wood particle in pastes and composites was investigated. Wood particles of fineness 0.42 mm and 1.20 mm were employed. Natural wood particles and the treated ones in cold or hot water (80 °C) were performed. The effects of the extractives solutions from the treatments applied to the wood particles on wood-gypsum compatibility were studied. For pastes and composites, water-to-gypsum ratio was 0.65. Wood particles-to-gypsum plaster ratios were 5%, 10% and 15%, in mass. Kinetics of temperature, mechanical performance and dynamic elasticity modulus by ultrasound measurements were applied to evaluate the gypsum plaster pastes and its composites behaviors. Results show that the extractive solutions changed the time of gypsum plaster hydration, being more sensitive to hot water treatment. The composites compressive strength increase with the wood particles pretreatment. The best result was to room temperature treatment. The same performance was found to the modulus of elasticity. Treatments have improved significantly the flexural strength. The best wood particle content was 10%. These results show the possibility of using this wood waste with an easy and simple treatment to make eco-efficient building materials.</span>
