Answer:
0.00335 moles
Explanation:
From the question, Using
PV = nRT................... Equation 1
Where P = pressure, V = Volume, n = number of moles of argon gas, R = Molar gas constant, T = Temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: P = 1 atm (standard pressure), T = 273 K (standard temperature), V = 75 mL = 0.075 dm³
Constant: R = 0.082 atm·dm³/K·mol
Substitute into equation 2
n = (1×0.075)/(273×0.082)
n = 0.075/22.386
n = 0.00335 moles
Answer:
First, place no. 5 in front of the CO2 in order to balance the carbon atoms. Next, place no. 6 in front of H2O to balance the hydrogen atoms. Lastly place no. 8 in front of the O2 so that there are 16 oxygen atoms on both sides of the reaction.
Answer:
9.9652g of water
Explanation:
The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:
n = PV / RT
Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)
Replacing:
n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K
n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:
1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>
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As the initial mass of water was 10g, the mass of water that remains in liquid phase is:
10g - 0.0348g = <em>9.9652g of water</em>
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I hope it helps!
Answer:
B = basic
Explanation:
Given data:
[OH⁻] = 5.35×10⁻⁴M
pH = ?
Solution:
pOH = -log[OH⁻]
pOH = - [5.35×10⁻⁴]
pOH = 3.272
it is known that,
pH + pOH = 14
pH = 14- pOH
pH = 14 - 3.272
pH = 10.728
The acidic pH is range from zero to less than 7 while 7 pH is neutral and above 7 the pH is basic. So, the given solution is basic.
