Answer:
<em>They could see how the spaceship is going to launch and how the trip is going to end up. I hope this helped...!</em>
Explanation:
<u>Chemistry</u> can be used for lots of things, for example, <em>engineering</em> and <em>science</em> such as <em>kinetic energy</em> or <em>mechanical energy</em>.
Answer:
Carbon
Carbon atoms have 6 protons.
<u>Answer:</u> The increase in pressure is 0.003 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure = ?
= Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![801^oC=[801+273]K=1074K](https://tex.z-dn.net/?f=801%5EoC%3D%5B801%2B273%5DK%3D1074K)
= final temperature = ![(801+1.00)^oC=802.00=[802+273]K=1075K](https://tex.z-dn.net/?f=%28801%2B1.00%29%5EoC%3D802.00%3D%5B802%2B273%5DK%3D1075K)
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28800J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B1074%7D-%5Cfrac%7B1%7D%7B1075%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D3%5Ctimes%2010%5E%7B-3%7Datm%5C%5C%5C%5CP_2%3De%5E%7B3%5Ctimes%2010%5E%7B-3%7D%7D%3D1.003atm)
Change in pressure = 
Hence, the increase in pressure is 0.003 atm
Answer:
8.28 MPa
Explanation:
From the question given above, the following data were obtained:
Radius (r) = 2×10¯³ m
Force applied (F) = 104 N
Pressure (P) =?
Next, we shall determine the area of the nail (i.e circle). This can be obtained as follow:
Radius (r) = 2×10¯³ m
Area (A) of circle =?
Pi (π) = 3.14
A = πr²
A = 3.14 × (2×10¯³)²
A = 3.14 × 4×10¯⁶
A = 1.256×10¯⁵ m²
Next, we shall determine the pressure. This can be obtained as follow:
Force applied (F) = 104 N
Area (A) = 1.256×10¯⁵ m²
Pressure (P) =?
P = F / A
P = 104 / 1.256×10¯⁵
P = 8280254.78 Nm¯²
Finally, we shall convert 8280254.78 Nm¯² to MPa. This can be obtained as follow:
1 Nm¯² = 1×10¯⁶ MPa
Therefore,
8280254.78 Nm¯² = 8280254.78 Nm¯² × 1×10¯⁶ MPa / 1 Nm¯²
8280254.78 Nm¯² = 8.28 MPa
Thus, the pressure exerted on the wall is 8.28 MPa
