Answer:
Slime mold or slime mould is an informal name given to several kinds of unrelated eukaryotic organisms that can live freely as single cells, but can aggregate together to form multicellular reproductive structures. Slime molds were formerly classified as fungi but are no longer considered part of that kingdom.
Answer:
The density of acetic acid at 30°C = 1.0354_g/mL
Explanation:
specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)
Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)
= 0.9956 g/mL × 1.040
= 1.0354_g/mL
Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.
Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure
Answer:
249 L
Explanation:
Step 1: Write the balanced equation
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈
The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol
Step 3: Convert "30.0°C" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 30.0°C + 273.15 = 303.2 K
Step 4: Calculate the volume of carbon dioxide
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm
V = 249 L
Answer:
5.06atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
V1 = initial volume (Litres)
V2 = final volume (Litres)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 1.34 atm
P2 = ?
V1 = 5.48 L
V2 = 1.32 L
T1 = 61 °C = 61 + 273 = 334K
T2 = 31 °C = 31 + 273 = 304K
Using P1V1/T1 = P2V2/T2
1.34 × 5.48/334 = P2 × 1.32/304
7.34/334 = 1.32P2/304
Cross multiply
334 × 1.32P2 = 304 × 7.34
440.88P2 = 2231.36
P2 = 2231.36/440.88
P2 = 5.06
The final pressure is 5.06atm
