Answer:
The total pressure after one half is 6.375 atm.
Explanation:
The initial pressure of product is increases while the pressure of reactant would decrease.
Balanced chemical equation:
2N₂O → 2N₂ + O₂
The pressure of N₂O is 5.10 atm. The change in pressure would be,
N₂O = -2x
N₂ = +2x
O₂ = +x
The total pressure will be
P(total) = P(N₂O) + P(N₂) + P(O₂)
P(total) = ( 5.10 - 2x) + (2x) + (x)
P(total) = 5.10 + x
After one half life:
P(N₂O) = 1/2(5.10) = 5.10 - 2x
x = 5.10 - 1/2(5.10) /2
x = 5.10 - 0.5 (5.10) /2
x = 5.10 - 2.55 / 2
x = 2.55 /2 = 1.275 atm
Thus the total pressure will be,
P(total) = 5.10 + x
P(total) = 5.10 + 1.275
P(total) = 6.375 atm
Answer:
The diagram is attached.
Explanation:
Peptide just refers to two or more amino acids liking together. A dipeptide consists of two amino acids linking together. A polypeptide is more that two amino acids linking together.
Answer:
First, we can test Solution 1. We know that Sodium Hydroxide is a strong base. If we test acids on blue litmus paper, they will turn red. If we test bases on red litmus paper, they will turn blue. So, you can test all the of the solutions- water, sodium hydroxide and hydrochloric acid with blue and red litmus paper. HCl, Hydrochloric acid is an acid, so it will turn blue litmus paper red. It will not turn red litmus blue. The acids will turn blue litmus paper red. The bases will turn red litmus paper blue. Only water is a neutral liquid, which will not turn blue litmus paper red or red litmus paper blue. It will not change the colour of it. Thus, if you test all the solutions with blue and red litmus paper, you will know which solution is water. Water is the only one which is neutral. It is the only solution which cannot change the colour of any litmus paper. Thus, you can identify it very easily.
The last question depends on the scientific method.
Scientific Method
1. Make an observation of what u are testing 2. Form a question about ur observations. 3. Make a hypothesis 4. Conduct your experiment 5. Record information and results.
The first question depends what you will need to use first for the experiment.
Answer:
The partial pressure of argon in the jar is 0.944 kilopascal.
Explanation:
Step 1: Data given
Volume of the jar of air = 25.0 L
Number of moles argon = 0.0104 moles
Temperature = 273 K
Step 2: Calculate the pressure of argon with the ideal gas law
p*V = nRT
p = (nRT)/V
⇒ with n = the number of moles of argon = 0.0104 moles
⇒ with R = the gas constant = 0.0821 L*atm/mol*K
⇒ with T = the temperature = 273 K
⇒ with V = the volume of the jar = 25.0 L
p = (0.0104 * 0.0821 * 273)/25.0
p = 0.00932 atm
1 atm =101.3 kPa
0.00932 atm = 101.3 * 0.00932 = 0.944 kPa
The partial pressure of argon in the jar is 0.944 kilopascal.
