Answer:
Explanation:
- The chemical reaction for cellular respiration is as follows -
C₆H₁₂O₆ ( Glucose ) + 6 O₂ ( Oxygen ) --------------> 6CO₂ ( Carbon dioxide ) + 6H₂O ( Water ) + 38 ATP molecules ( Energy ) .
This is an example of oxidation of carbon .
- The chemical reaction for photosynthesis is as follows -
6CO₂ ( Carbon dioxide ) + 6H₂O ( Water ) --------------> C₆H₁₂O₆( Glucose ) + 6 O₂ ( Oxygen ).
This is an example of reduction of carbon .
Answer:
<h2>
<em>no</em></h2>
Explanation:
<h2><u><em>
the particles in gas move so freely that it cannot have a definite density</em></u></h2><h2><u><em>
</em></u></h2><h2><u><em>
</em></u></h2><h2><u><em>
</em></u></h2><h2><u><em>
moo</em></u></h2>
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
You need to find moles of the gas, so you would use the ideal gas law:
PV=nRT
Pressure
Volume
n=moles
R= gas constant
Tenperature in Kelvin
n= PV/RT
(1.00atm)(1.35L)/(.08206)(332K) = 0.050mol
Molar mass is grams per mole, so
(3.75g/.050mol) = 75g/mol
Answer:
True
Explanation:
In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].
The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.
This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.
Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.
