False
Explanation:
The best way to ensure scientific claims are valid is to subject them to various tests. Scientific claims has nothing to do with the body presenting them. Claims are authenticated following due scientific process.
- Scientific method provides a way of ascertaining the validity of an assumption.
- If the claims is not supported by the body of evidence obtained, then the claim has to be modified or refuted.
- When the claim is consistent with various infallible proof from different sources, then it can be accepted.
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Answer:
you are right
Explanation:
and i do not need to explain it because you did
Acetonitrile refers to the chemical compound exhibiting the formula CH₃CN. It is the basic organic nitrile and a colorless liquid. It is generated primarily as a byproduct of acrylonitrile production. It is used as a polar aprotic solvent in the purification of butadiene and in organic synthesis.
The addition of water to cyanides generates amines and acids. Acetonitrile produces ammonia and acetic acid.
CH₃CN (acetonitrile) → CH₃COOH (acetic acid) + NH₃ (ammonia)
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
