The ideal gas law may be written as
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore
Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)
Explanation:
the conductors are the three u have checked
Answer : The correct option is A.
Explanation :
Enzyme-catalyzed reaction :
Enzyme act as a biological catalyst and the role of catalyst is to increase the rate of chemical reaction by lowering the activation energy.
Most of the chemical reactions are slow in the absence of enzyme but in the presence of enzyme, the reaction become faster. That means the Enzyme accelerate the rate of reaction.
Therefore, the correct answer is the reaction is faster than the same reaction in the absence of the enzyme.
Here the base is a benzoate ion, which is a weak base and reacts with water.
The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.
Therefore [OH-] = [C6H5COOH]
In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]
pOH = 14 - pH
pH given = 9.04
pOH = 14-9.04 = 4.96
pOH = -log[OH-] or ![[OH^{-}] = 10^{^{-pOH}}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%2010%5E%7B%5E%7B-pOH%7D%7D%20)
![[OH^{-}] = 10^{^{-4.96}}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%2010%5E%7B%5E%7B-4.96%7D%7D%20)
![[OH^{-}] = 1.1\times 10^{-5}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%201.1%5Ctimes%2010%5E%7B-5%7D%20)
The base dissociation equation kb = 
![kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}](https://tex.z-dn.net/?f=%20kb%20%3D%5Cfrac%7B%5BC6H5COOH%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BC6H5COO%5E%7B-%7D%5D%7D)
H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.
Value of Kb is given = 
And value of [OH-] we have calculated as 
and value of C6H5COOH is equal to OH-
Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-
![1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}](https://tex.z-dn.net/?f=%201.6%5Ctimes%2010%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5BC6H5COO%5E%7B-%7D%5D%7D%20)
![[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}](https://tex.z-dn.net/?f=%20%5BC6H5COO%5E%7B-%7D%5D%20%3D%20%5Cfrac%7B%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B1.6%5Ctimes%2010%5E%7B-10%7D%7D%20)
![[C6H5COO^{-}] = 0.76 M](https://tex.z-dn.net/?f=%20%5BC6H5COO%5E%7B-%7D%5D%20%3D%200.76%20M%20)
or 
So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L
Moles of NaC6H5COO would be = 
Moles of NaC6H5COO (sodium benzoate) = 0.38 mol
Answer:
0.48
Explanation:
because we are looking for I which is current we say
I = 12÷25 which is 0.48.
