Here, we are going to calculate the mass % of C in the mixture.
What is a Mixture?
A mixture is composed of one or more pure substances in varying composition. There are two types of mixtures: heterogeneous and homogeneous. Heterogeneous mixtures have visually distinguishable components, while homogeneous mixtures appear uniform throughout.
Given that,
The mass % of CO =35.0% =35.0 g in 100 g mixture
The mass % of CO2 = 65% =65 g in 100 g mixture
Therefore,
The mass of C from CO = 15.007 g C
Similarly,
The mass of C from CO2 = 17.738 g C
Thus, the total mass of C = 15.007 g+17.738 g =32.745 g
Therefore,
The mass % of C= 32.745% =32.7%
Thus, the mass % of C in the mixture is 32.7%
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Answer:
The second experiment (reversible path) does more work
Explanation:
Step 1:
A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C
<em>(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm</em>
<em></em>
Irreversible path: w =-Pex*ΔV
⇒ with Pex = 1.00 atm
⇒ with ΔV = 1.20 L
W = -(1.00 atm) * 1.20 L
W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J
<em>(b) The gas is allowed to expand reversibly and isothermally to the same final volume.</em>
<em></em>
W = -nRTln(Vfinal/Vinitial)
⇒ with n = the number of moles = 0.200
⇒ with R = gas constant = 8.3145 J/K*mol
⇒ with T = 298 Kelvin
⇒ with Vfinal/Vinitial = 2.40/1.20 = 2
W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)
W = -343.5 J
The second experiment (reversible path) does more work
<span>KE = 1/2mv^2
KE = 1/2(8)4 m/s^2
KE = 4*4
KE = 16 Joules
Kinetic energy would equal 16 J </span>
<span>A)photosynthetic bacteria</span>
