Answer: 91.73g of NaCl
Explanation:
First, we solve for the number of moles of F2 using the ideal gas equation
V = 12L
P = 1.5 atm
T = 280K
R = 0.082atm.L/mol/K
n =?
PV = nRT
n = PV /RT
n = (1.5x12)/(0.082x280)
n = 0.784mol
Next, we convert this mole ( i.e 0.784mol) of F2 to mass
MM of F2 = 19x2 = 38g/mol
Mass conc of F2 = n x MM
= 0.784 x 38 = 29.792g
Equation for the reaction is given below
F2 + 2NaCl —> 2NaF + Cl2
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass conc. of NaCl from the equation = 2 x 58.5 = 117g
Next, we find the mass of NaCl that reacted with 29.792g of F2.
From the equation,
38g of F2 redacted with 117g of NaCl.
Therefore, 29.792g of F2 will react with Xg of NaCl i.e
Xg of NaCl = (29.792 x 117)/38
= 91.73g
Therefore, 91.73g of NaCl reacted with f2
Answer:
We have to add 2.30 L of oxygen gas
Explanation:
Step 1: Data given
Initial volume = 4.00 L
Number of moles oxygen gas= 0.864 moles
Temperature = constant
Number of moles of oxygen gas increased to 1.36 moles
Step 2: Calculate new volume
V1/n1 = V2/n2
⇒V1 = the initial volume of the vessel = 4.00 L
⇒n1 = the initial number of moles oxygen gas = 0.864 moles
⇒V2 = the nex volume of the vessel
⇒n2 = the increased number of moles oxygen gas = 1.36 moles
4.00L / 0.864 moles = V2 / 1.36 moles
V2 = 6.30 L
The new volume is 6.30 L
Step 3: Calculate the amount of oxygen gas we have to add
6.30 - 4.00 = 2.30 L
We have to add 2.30 L of oxygen gas
Answer:
That the enzyme is a protein
Explanation:
Remember the proteins are groups of thousands of amino acids, by their owns, since they are very small units, they can not act as a catalyst.
But as a polymer, the protein enzyme, have different shape, size and physical and chemical properties than a single monomer.
Remember also the proteins to be active, they need certain number of amino acids join together to form a specific shape that is going to match with another molecule to speed the chemical reaction and act as an enzyme.
<u>Answer:</u> The standard enthalpy of formation of 
is -92.7 kJ/mol
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_3O_2(g))})]-[(2\times \Delta H^o_f_{(CO(g))})+(1\times \Delta H^o_f_{(C(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_3O_2%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C%28s%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![127.3=[(1\times \Delta H^o_f_{(C_3O_2(g))})]-[(2\times (-110))+(1\times (0))]\\\\\Delta H^o_f_{(C_3O_2(g))}=-92.7kJ/mol](https://tex.z-dn.net/?f=127.3%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28C_3O_2%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%28-110%29%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28C_3O_2%28g%29%29%7D%3D-92.7kJ%2Fmol)
Hence, the standard enthalpy of formation of is -92.7 kJ/mol
