Answer:
B. a chemical reactions that occurs in the chloroplast of a plant.
Explanation:
HCl and NaOH react in a 1:1 ratio, meaning that 1 H+ from HCl will react with 1 OH- from NaOH. Knowing this, and that molarity is mol/liter, all we need to do is use what we have available. First we must find the mols of HCl in our solution, so we set up the following equation in the following steps:
1. 24.75mL x (0.359mol NaOH / 1000mL) = 8.885 x 10^-3mol NaOH
This is done in order to find the mols of NaOH to convert to mols of HCl.
2. 8.885x10^-3mol NaOH x (1 mol HCl/1mol NaOH) = 8.885 x 10^-3mol HCl
Here we just used the mols of NaOH we found to convert to mols of HCl using the 1:1 ratio described earlier.
From the mols of HCl all we have to do is divide by the amount of liters in the solution. Since we started with 10mL HCl and added 24.75mL NaOH, the total volume is 34.75mL = 0.03475L. So:
8.885 x 10^-3mol HCl/0.03475L = 2.557 x 10^-1M HCl
However, this is the molarity of the HCl and NaOH solution, not the original HCl solution. Using the dilution equation M1V1=M2V2, we can solve for the original molarity.
M1 = the molarity of our HCl in the titrated mixture (2.557 x 10^-1M HCl)
V1 = the total volume that our mixture has (34.75mL = 0.03475L)
M2 = what we're trying to find
V2 = the amount of the original HCl that we had (10mL = 0.010L)
Simply solving for M2 gives us:
M2 = (M1V1) / V2 or:
M2=((2.557 x 10^-1) x 0.03475L) / 0.010L = 8.89 x 10^-1M HCl. That is your answer.
Answer:
a
b
c
Explanation:
From the question we are told that
The reaction of cyclobutane and oxygen is
ΔH°f (kJ mol-1) : C4H8(g) = 27.7 ; CO2(g) = -393.5 ; H2O(g) = -241.8 ΔH° = kJ
Generally ΔH° for this reaction is mathematically represented as
![\Delta H^o _{rxn} = [[4 * \Delta H^o_f (CO_2_{(g)} ) + 4 * \Delta H^o_f(H_2O_{(g)} ] -[\Delta H^o_f (C_2H_6_{(g)} + 6 * \Delta H^o_f (O_2_{(g)}) ] ]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%20_%7Brxn%7D%20%3D%20%5B%5B4%20%2A%20%5CDelta%20H%5Eo_f%20%28CO_2_%7B%28g%29%7D%20%29%20%2B%204%20%2A%20%5CDelta%20H%5Eo_f%28H_2O_%7B%28g%29%7D%20%5D%20-%5B%5CDelta%20H%5Eo_f%20%28C_2H_6_%7B%28g%29%7D%20%2B%206%20%2A%20%5CDelta%20H%5Eo_f%20%28O_2_%7B%28g%29%7D%29%20%5D%20%5D)
=> ![\Delta H^o _{rxn} = [[4 * (-393.5) + 4 * (-241.8) ] -[ 27.7 + 6 * 0]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%20_%7Brxn%7D%20%3D%20%5B%5B4%20%2A%20%28-393.5%29%20%2B%204%20%2A%20%28-241.8%29%20%5D%20-%5B%2027.7%20%2B%206%20%2A%200%5D)
=>
Generally the total heat capacity of 4 mol of CO2(g) and 4 mol of H2O(g), using CCO2(g) = 37.1 J K-1 mol-1 and CH2O(g) = 33.6 J K-1 mol-1. C = J K-1 is mathematically represented as
![H = [ 4 * C_{CO_2_{(g)}} + 6* C_{CH_2O_{(g)}}]](https://tex.z-dn.net/?f=H%20%20%3D%20%5B%204%20%2A%20C_%7BCO_2_%7B%28g%29%7D%7D%20%2B%206%2A%20C_%7BCH_2O_%7B%28g%29%7D%7D%5D)
=> ![H = [ 4 * 37.1 + 6* 33.6 ]](https://tex.z-dn.net/?f=H%20%20%3D%20%5B%204%20%2A%2037.1%20%2B%206%2A%2033.6%20%5D)
=>
From the question the initial temperature of reactant is 
Generally the enthalpy change(
) of the reaction is mathematically represented as
=> 
=>
The balanced net ionic equation for the reaction of aqueous sodium sulfide with aqueous lead(II) nitrate is Pb2+(aq) + S2-(aq) ===> PbS(s).
The chemical equation known as the net ionic equation only displays the substances that are directly involved in the chemical process. It is described how to write a net ionic equation. We only need to take out the spectator ions in order to formulate the net ionic equation. We are left with the net ionic equation after removing the spectator ions! The net ionic equation can be obtained by repeating this procedure for any reaction.
Na2S(aq) + Pb(NO3)2(aq) ===> 2NaNO3(aq) + PbS(s) ... molecular equation
sodium sulfide....lead(II) nitrate .........sodium nitrate.....lead sulfide
2Na+(aq) + S2-(aq) + Pb2+(aq) + 2NO3-(aq) ==> 2Na+(aq) + 2NO3-(aq) + PbS(s) ... complete ionic equation
Pb2+(aq) + S2-(aq) ===> PbS(s) ... net ionic equation
To know more about net ionic equation refer to: brainly.com/question/22885959
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