Answer: The molecular formula will be 
Explanation:
When percentages are provided, we assume the total mass to be 100 grams.
Thus, the mass of each element corresponds to the specified percentage.
Mass of C= 70.6 g
Mass of H = 5.9 g
Mass of O = 23.5 g
Step 1: convert given masses to moles.
Moles of C =
Moles of H =
Moles of O =
Step 2: For determining the mole ratio, divide each molar amount by the smallest number of moles calculated.
For C = 
For H =
For O =
The resulting ratio of C: H: O= 4: 4: 1
Hence, the empirical formula obtained is 
The empirical weight is calculated as = 4(12)+4(1)+1(16)= 68g.
The molecular weight = 136 g/mole
Now the molecular formula needs to be obtained.
The molecular formula can be derived as=
Answer:
Refer to the explanation.
Explanation:
Formation reactions involve the creation of one mole of a compound from its elements in their standard states.
NaBr (s)
The equation for the standard formation is
Na (s) + (1/2)Br₂ (g) → NaBr (s)
As per appendix C, the standard heat of formation for NaBr(s) is
ΔH∘f = -359.8 kJ/mol.
SO₃ (g)
The equation for the standard formation is
S (s) + (3/2) O₂ (g) → SO₃ (g)
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ΔH∘f = -395.2 kJ/mol.
Pb(NO₃)₂ (s)
The equation for the standard formation is
Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)
According to appendix C, the standard heat of formation for Pb(NO₃)₂(s) is
ΔH∘f = -451.9 kJ/mol.
I hope this is helpful!
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To calculate the moles of MgSO4.7H2O, we find the molar mass equals 246, thus moles = 32 / 246 = 0.13 moles. Upon heating, all 7 H2O from one molecule will evaporate. The total moles of H2O present amount to 7 x 0.13 = 0.91, and the mass of that H2O is 0.91 x 18 = 16.38g. Therefore, the mass of the anhydrous MgSO4 that remains is 32 - 16.38 = 15.62 g.
Answer:
B,C,D
Explanation:
The quantity of CCl4 produced is contingent on the amount of CH4 used in a 1:1 ratio. Given that there are twice as many moles of Cl2 compared to CH4, some Cl2 will remain unreacted. To fully utilize all Cl2, additional CH4 must be introduced into the reaction.
The accurate answer is option 2. When Pt2+ undergoes reduction, it loses electrons, thus oxidizing itself. The oxidation state corresponds to the atom's charge. When an electron is added, the overall charge diminishes, which in turn reduces the oxidation number.
