The pressure of the tyre after the temperature change is 0.89atm. Details about pressure can be found below.
<h3>How to calculate pressure?</h3>
The pressure of the tyre can be calculated using the following equation:
P1/T1 = P2/T2
Where;
- P1 = initial pressure
- P2 = final pressure
- T1 = initial temperature
- T2 = final temperature
According to this question, a tyre at 21°C (294K) has a pressure of 0.82 atm. Its temperature decreases to –3.5°C (269.5K).
0.82 × 294 = P2 × 269.5
241.08 = 269.5P2
P2 = 241.08 ÷ 269.5
P2 = 0.895atm
Therefore, the pressure of the tyre after the temperature change is 0.89atm.
Learn more about pressure at: brainly.com/question/23358029
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Answer:
The value of 
of the an ethylamine is 
.
Explanation:
The pH of the solution = 12.067
The pOH of the solution = 14 - pH =14-12.607 =1.933
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1.933=-\log[OH^-]](https://tex.z-dn.net/?f=1.933%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.0117 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.0117%20M)
Initially
0.342 M 0 0
At equilibrium
(0.342-x) x x
The value of x =
The expression of is given as:
![K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BC_2H_5NH_3%5E%7B%2B%7D%5D%5BOH%5E-%5D%7D%7B%5BC_2H_5NH_2%5D%7D)
The value of of the an ethylamine is .
Answer:
1.32*10^23 molecules
Explanation:
sucrose formula: C12H22O11
molar mass: 12(12.01)+22(1.01)+11(16.00)=342.34g/mol
75.0 g C12H22O11 * (1 mol C12H22O11)/(342.34g C12H22O11)=0.219 mol C12H22O11
0.219 mol * (6.022*10^23)/mol = 1.32*10^23 molecules (three sig. figures)
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value. The equation to describe the decay is
Nt=N0(1/2)
where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time. So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
