Answer:
A chemical formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule, using chemical element symbols, numbers, and sometimes also other symbols, such as parentheses, dashes, brackets, commas and plus (+) and minus (−) signs.
Compounds used as medicines are most often organic compounds, which are often divided into the broad classes of small organic molecules (e.g., atorvastatin, fluticasone, clopidogrel) and "biologics" (infliximab, erythropoietin, insulin glargine), the latter of which are most often medicinal preparations of proteins
Explanation:
Answer:
Mass = 5.56 g
Explanation:
Given data:
Mass of Cl₂ = 4.45 g
Mass of NaCl produced = ?
Solution:
Chemical equation:
2Cl₂ + 4NaOH → 3NaCl + NaClO₂ + 2H₂O
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 4.45 g/ 71 g/mol
Number of moles = 0.063 mol
Now we will compare the moles of Cl₂ with NaCl.
Cl₂ : NaCl
2 : 3
0.063 : 3/2×0.063 =0.095 mol
Mass of NaCl:
Mass = number of moles × molar mass
Mass = 0.095 mol × 58.5 g/mol
Mass = 5.56 g
Answer:
-238.54 kJ/mol.
Explanation:
- We need to calculate the standard enthalpy of formation of liquid methanol, CH₃OH(l) that has the equation:
C(graphite) + 2H₂(g) + ½ O₂(g) → CH3OH(l) ΔHf° = ?
?? kJ/mol.
- using the information of the three equations:
(1) C(graphite) + O₂(g) → CO₂(g), ΔHf₁° = -393.5 kJ/mol
.
(2) H2(g) + ½ O₂(g) → H₂O(l), ΔHf₂° = -285.8 kJ/mol
.
(3) CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l), ΔH₃° = -726.56 kJ/mol
.
- We rearranging and add equations (1, 2, and 3) in such a way as to end up with the needed equation:
- equation (1) be as it is:
(1) C(graphite) + O₂(g) → CO₂(g), ΔHf₁° = -393.5 kJ/mol
.
- equation (2) should be multiplied by (2) and also the value of ΔHf₂°:
(2) 2H2(g) + O₂(g) → 2H₂O(l), ΔHf₂° = 2x(-285.8 kJ/mol
).
- equation (3) should be reversed and also the value of ΔH₃° should be multiplied by (-1):
(3) CO₂(g) + 2H₂O(l) → CH₃OH(l) + 3/2 O₂(g), ΔH₃° = 726.56 kJ/mol
.
- By summing the modified equations, we can get the needed equation and so:
The standard enthalpy of formation of liquid methanol, CH₃OH(l) = ΔHf₁° + 2(ΔHf₂°) - ΔH₃° = (-393.5 kJ/mol
) + 2(-285.8 kJ/mol
) - (- 726.56 kJ/mol) = -238.54 kJ/mol.