Answer:
the mole fraction of Gas B is xB= 0.612 (61.2%)
Explanation:
Assuming ideal gas behaviour of A and B, then
pA*V=nA*R*T
pB*V=nB*R*T
where
V= volume = 10 L
T= temperature= 25°C= 298 K
pA and pB= partial pressures of A and B respectively = 5 atm and 7.89 atm
R= ideal gas constant = 0.082 atm*L/(mol*K)
therefore
nA= (pA*V)/(R*T) = 5 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 2.04 mole
nB= (pB*V)/(R*T) = 7.89 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 3.22 mole
therefore the total number of moles is
n = nA +nB= 2.04 mole + 3.22 mole = 5.26 mole
the mole fraction of Gas B is then
xB= nB/n= 3.22 mole/5.26 mole = 0.612
xB= 0.612
Note
another way to obtain it is through Dalton's law
P=pB*xB , P = pA+pB → xB = pB/(pA+pB) = 7.69 atm/( 5 atm + 7.89 atm) = 0.612
Answer:
70.0 %
Explanation:
Step 1: Given data
- Mass of nitrogen (mN): 74.66 g
- Mass of the compound (mNxOy): 250 g
Step 2: Calculate the mass of oxygen (mO) in the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
mNxOy = mN + mO
mO = mNxOy - mN
mO = 250 g - 74.66 g = 175 g
Step 3: Determine the percent composition of oxygen in the sample
We will use the following expression.
%O = mO / mNxOy × 100%
%O = 175 g / 250 g × 100% = 70.0 %
Answer:
D (or E If properly listed to include the active site option)
Explanation:
A. Is true
Enzymes are organically biochemical catalyst and thus they can speed up the rate of chemical reaction in the body
B is true
They are catalysts as said earlier
C is true
They have active sites. An enzyme does not act on all substrates. They have particular group on which they can act. For example, we have carbohydrates enzymes that act on carbohydrates substrate only. This enzymes have no business acting on a protein substrate.
D. Enzymes are proteins
One of the important characteristics of enzymes is that they are protenious in nature
E. This is wrong. Enzymes like any over catalyst are not consumed in the course of the biochemical reaction
Answer: option b. 1-chloro-4,4-dimethyl-2-pentene
Explanation: the numbering is done from the side that gives the double bond the lowest low count
