Answer:
option c
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Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
Answer:
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Explanation:
The solubility product constant, Ksp, is calculated by the product of the concentration of the dissociated ions raise to the stoichiometric coefficient of each ion. We calculate what is asked in this problem as follows:
Ag2SO4 = 2Ag^+ + (SO4)^2-
Ksp = [Ag^+]^2 [SO4^2-]
<span>6.0 x10^-5 = 4x^2 (x) = 4x^3
x = 0.025 M
0.025 mol </span>(SO4)^2- / L ( 1 mol Ag2SO4 / 1 mol (SO4)^2-) (311.799 g / mol ) = 7.69 g Ag2SO4 / L solution
Answer : The final volume of the gas is, 34.8 L
Explanation :
Formula used :

where,
w = work done = 137.1 J = 35.8 L.atm (1 L.atm = 101.3 J)
p = external pressure = 783 torr = 1.03 atm (1 atm = 760 torr)
= final volume = ?
= initial volume = 64.0 mL = 0.0640 L (1 L = 1000 mL)
Now put all the given values in the above formula, we get:


Thus, the final volume of the gas is, 34.8 L