Explanation:
Medicinal chemistry and pharmaceutical chemistry are disciplines at the intersection ofchemistry, especially synthetic organicchemistry, and pharmacology and various other biological specialties, where they are involved with design,chemical synthesis and development for market of pharmaceutical agents, or bio-active ...
Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
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0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
To learn more about stoichiometric quantities visit:
<h3>
brainly.com/question/28174111</h3>
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Answer:
The number of carbon atoms in the container is 1.806 × 10²⁴ or the container contains 1.806 × 10²⁴ atoms of carbon
Explanation:
By Avogadro's number, 1 mole of a substance contains 6.02 × 10²³ particles of the substance
Here we have 0.45 mole of CO₂ contains
0.45 × 6.02 × 10²³ particles of CO₂ that is 2.709 × 10²³ particles of CO₂ or equivalent to 2.709 × 10²³ atoms of Carbon
Similarly, 2.55 moles of CaC₂ contains 2.55 × 6.02 × 10²³ particles of CaC₂ or 1.5351 × 10²⁴ atoms of Carbon
The total number of carbon atoms is therefore;
2.709 × 10²³ + 1.5351 × 10²⁴ = 1.806 × 10²⁴ atoms of carbon.
