Answer:
4.81 moles
Explanation:
The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.
Pressure at which gauge reads zero = 14.7 psi
Pressure read by the gauge = 988 psi
Total pressure = 14.7 + 988 psi = 1002.7 psi
Also, P (psi) = P (atm) / 14.696
Pressure = 1002.7 / 14.696 = 68.2297 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 4.81 moles
The number of molecules decrease
The answer to this question is False
Answer:
have stars that might appear to wobble
often have one star that is brighter than the other
Explanation:
A binary star system is a star system made up of mostly two stars that moves round their common fixed center.
The two orbiting stars are gravitationally bonded to one another and they move round each other.
Most binary stars might appear wobble. One of the stars often appears brighter than the other.
Answer:

Explanation:
Hello!
In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:
![[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}](https://tex.z-dn.net/?f=%5BMgCl_2%5D%3D1.672%5Cfrac%7BmolMgCl_2%7D%7BL%7D%2A%5Cfrac%7B95.211gMgCl_2%7D%7B1molMgCl_2%7D%3D159.2%5Cfrac%7BgMgCl_2%7D%7BL%7D)
Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:
![[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14](https://tex.z-dn.net/?f=%5BMgCl_2%5D%3D159.2%5Cfrac%7BgMgCl_2%7D%7BL%7D%2A%5Cfrac%7B1L%7D%7B1000mL%7D%2A%5Cfrac%7B1mL%7D%7B1.137g%7D%3D0.14)
Which is also the by-mass fraction and in percent it turns out:

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