The given 2.6 µC of charge is due to a buildup of electrons, each of which has a charge of 1.6 x 10^-19 C. The 2.6 <span>µC is equivalent to 2.6 x 10^-6 C, so we can divide this by the individual charge of an electron:
</span>2.6 x 10^-6 C / 1.6 x 10^-19 (C/electron) = 1.625 x 10^13 electrons
Answer:
Enzyme Active Site and Substrate Specificity
There may be one or more substrates for each type of enzyme, depending on the particular chemical reaction. In some reactions, a single-reactant substrate is broken down into multiple products. In others, two substrates may come together to create one larger molecule.
The equation for work is
W = PdV
and it is integrated and limits are the conditions of state 1 and state 2
If the gas is ideal and the expansion is isothermal, then P = nRT/V and the equation can be integrated with respect to V
If the process is adiabatic, the equation P1V1^g = PV^g can be used to substitue P in terms of conditions of State 1.
Answer:
28.28 L.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and T are constant, and have two different values of V and P:
<em>P₁V₁ = P₂V₂</em>
<em></em>
P₁ = 700.0 mm Hg, V₁ = 4.0 L.
at burst: P₂ = 99.0 mm Hg, V₂ = ??? L.
<em>∴ V₂ = P₁V₁/P₂</em> = (700.0 mm Hg)(4.0 L)/(99.0 mm Hg) = <em>28.28 L.</em>
