Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
Answer: The concentration of 
is 0.234 M
Explanation:
According to the neutralization law,
where,
= basicity = 2
= molarity of solution = ?
= volume of solution = 50.0 ml
= acidity of 
= 1
= molarity of solution = 0.375 M
= volume of solution = 62.5 ml
Putting in the values we get:
Therefore concentration of is 0.234 M
Answer:
composting scraps
recycling is the action or process of converting waste into reusable material.
Answer:
Deletion
Explanation:
A mutation refers to a change in the DNA sequence. This means that the original sequence of bases in the DNA has been altered permanently.
There are three types of DNA mutations; base substitutions, deletions and insertions.
In the particular case of this question, the original is TTCGATA while the copy is TTGATA. If you look closely at the two, you will notice that the C has been omitted. This is an example of deletion.
Answer:
5.5 atm
Explanation:
Step 1: Calculate the moles in 2.0 L of oxygen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
2.0 L × 1 mol/22.4 L = 0.089 mol
Step 2: Calculate the moles in 8.0 L of nitrogen at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
8.0 L × 1 mol/22.4 L = 0.36 mol
Step 3: Calculate the total number of moles of the mixture
n = 0.089 mol + 0.36 mol = 0.45 mol
Step 4: Calculate the pressure exerted by the mixture
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 0.45 mol × (0.0821 atm.L/mol.K) × 298 K / 2.0 L = 5.5 atm
