Answer:
Final Temperature = 36.54 ⁰C
Explanation:
Lets suppose the gas is acting ideally, then according to Charle's Law, "<em>The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature</em>". Mathematically for initial and final states the relation is as follow,
V₁ / T₁ = V₂ / T₂
Data Given;
V₁ = 32 L
T₁ = 10 °C = 283.15 K ∴ K = °C + 273.15
V₂ = 35 L
T₂ = ??
Solving equation for T₂,
T₂ = V₂ × T₁ / V₁
Putting values,
T₂ = (35 L × 283.15 K) ÷ 32 L
T₂ = 309.69 K ∴ ( 36.54 °C )
Result:
As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.
the correct IUPAC name of the compound is 1-Butanal.
<h3>What are IUPAC names?</h3>
It is a system of naming organic compounds based on the longest carbon-to-carbon single bonds. It does not matter whether these longest chains are continuous or in a ring.
Thus, when the compound with the chemical formula, CH3-CH2-CH2CHO is considered. The longest carbon-to-carbon chain is 4. The 1st carbon carries a functional group known as an aldehyde.
Aldehydes are equipped with the carbonyl group and have the general formula R−CH=O. They are also sometimes referred to as formyl.
Aldehydes are named after their parent alkane chains with a slight modification. The 'e' is replaced with 'al'
The aldehyde in this case has four carbons. This means that the parent alkane is Butane. Therefore, the name of the compound will be 1-Butanal.
More on IUPAC names can be found here: brainly.com/question/16631447
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C. A central body is the center of the universe.
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
