Answer:
V = 85.2
Explanation:
STP = 273K and 1 atm
Considering what we know about STP, we get the moles, temperature, and pressure. Using the ideal gas law we can find the volume (PV = nRT). Plug in our variables: (1 * V = 3.80 * R * 273). Since we are dealing with atm and not kPA or mmHg, we use the constant for atm (0.0821) which we use for R. (So.. now our equation is 1 * V = 3.80 * 0.0821 * 273). We now multiply the right side to get 85.17054. So... V = 85.2 considering sigificant figures (this is the part where I am the least sure of, since I havent done sig figs in a while)
The partial pressure of methane in the mixture of methane and ethane has been 1 atm.
Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.
The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.
The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.
However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.
For more information about the partial pressure, refer to the link:
brainly.com/question/14623719
Answer:
The answer to your question would be "the insect touching the trigger hairs".
Explanation:
I'm not much of an ex-plainer but I know this is the right answer because I took the test and got a 100%. Please trust me on this. If wrong, please tell me. This is what I was taught at school. Thank you and good day.
Answer:
Element
Explanation:
Because elements can be made with only one atom the rest of the answers cant be
Answer:
The final temperature of the solution is 44.8 °C
Explanation:
assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:
Q dis + Q sol = 0
Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C) is q dis= -83.3 KJ/mol . And the molecular weight is
M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol
Q dis = q dis * n = q dis * m/M = -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ
Qdis= -9.84 KJ
Also Qsol = ms * Cs * (T - Ti)
therefore
ms * Cs * (T - Ti) + Qdis = 0
T= Ti - Qdis * (ms * Cs )^-1 =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ
T= 44.8 °C
