I would love to help but I cant see the picture
Answer:
look at peaks:
sharp peak at ~ 1700 = carbonyl
broad peak at ~ 3300 = alcohol
both = mixture
Explanation:
In the first spectrum, the broad peak occurring in the range of 3300-3700 cm-1 corresponds to OH bond stretch frequency. Carbonyl C=O bond stretch has a frequency range of 1670-1820 cm-1. The absence of any peaks in this region in the given spectrum rules out the presence of carbonyl compound in the reaction mixture. So, the reaction mixture contains only pure alcohol.
The second spectrum shows a peak around 1700 cm-1 corresponding to the carbonyl compound along with the broad peak around 3300 cm-1 corresponding to alcohol. So, the reaction mixture contains both alcohol and carbonyl compounds.
A table containing different functional groups and their corresponding Infra Red frequencies is given below for further references
<u>Answer:</u> The enthalpy of formation of HCN(g) is 135.1 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_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
![2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)](https://tex.z-dn.net/?f=2NH_3%28g%29%2B3O_2%28g%29%2B2CH_4%28g%29%5Crightarrow%202HCN%28g%29%2B6H_2O%28g%29)
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN(g))})+(6\times \Delta H_f_{(H_2O(g))})]-[(2\times \Delta H_f_{(NH_3(g))})+(3\times \Delta H_f_{(O_2(g))})+(2\times \Delta H_f_{(CH_4(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%28g%29%29%7D%29%2B%286%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%28g%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H_f_%7B%28O_2%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28CH_4%28g%29%29%7D%29%5D)
We are given:
![\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-80.3kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.6kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H_f_%7B%28H_2O%28g%29%29%7D%3D-241.8kJ%2Fmol%5C%5C%5CDelta%20H_f_%7B%28NH_3%28g%29%29%7D%3D-80.3kJ%2Fmol%5C%5C%5CDelta%20H_f_%7B%28CH_4%28g%29%29%7D%3D-74.6kJ%2Fmol%5C%5C%5CDelta%20H_f_%7B%28O_2%28g%29%29%7D%3D0kJ%2Fmol%5C%5C%5CDelta%20H_%7Brxn%7D%3D-870.8kJ)
Putting values in above equation, we get:
![-870.8=[(2\times \Delta H_f_{(HCN(g))})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN(g))}=135.1kJ/mol](https://tex.z-dn.net/?f=-870.8%3D%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCN%28g%29%29%7D%29%2B%286%5Ctimes%20%28-241.8%29%29%5D-%5B%282%5Ctimes%20%28-80.3%29%29%2B%283%5Ctimes%20%280%29%29%2B%282%5Ctimes%20%28-74.6%29%29%5D%5C%5C%5C%5C%5CDelta%20H_f_%7B%28HCN%28g%29%29%7D%3D135.1kJ%2Fmol)
Hence, the enthalpy of formation of HCN(g) is 135.1 kJ/mol
The mass of oxygen required is 17.06 g; the mass of carbon dioxide produced is 300 g; the mass of copper (ii) nitrate produced is 16.7 g.
<h3>What is the equation of the synthesis of water from hydrogen and oxygen?</h3>
The equation of the synthesis of water from hydrogen and oxygen is given as follows:
The grams of oxygen required is given below:
![19.2 g\:of H_{2} \times\frac{1\:mol H_{2}}{18 g\:H_{2}} \times\frac{1mol\:O_{2}}{2 mol\:H_{2}}\times\frac{32g\:O_{2}}{1 mol\:O_{2}} = 17.06 g\: O_{2}](https://tex.z-dn.net/?f=19.2%20g%5C%3Aof%20H_%7B2%7D%20%5Ctimes%5Cfrac%7B1%5C%3Amol%20H_%7B2%7D%7D%7B18%20g%5C%3AH_%7B2%7D%7D%20%5Ctimes%5Cfrac%7B1mol%5C%3AO_%7B2%7D%7D%7B2%20mol%5C%3AH_%7B2%7D%7D%5Ctimes%5Cfrac%7B32g%5C%3AO_%7B2%7D%7D%7B1%20mol%5C%3AO_%7B2%7D%7D%20%3D%2017.06%20g%5C%3A%20O_%7B2%7D)
The equation of the combustion of propane is given as follows:
![C_{3}H_{8} + 5\:O_{2}\rightarrow3\:CO_{2}+4\:H_{2}O](https://tex.z-dn.net/?f=C_%7B3%7DH_%7B8%7D%20%2B%205%5C%3AO_%7B2%7D%5Crightarrow3%5C%3ACO_%7B2%7D%2B4%5C%3AH_%7B2%7DO)
The mass of carbon dioxide produced is given below:
![100 g\:of\:C_{3}H_{8} \times\frac{1\:mol\:C_{3}H_{8}}{44 g\:C_{3}H_{8}} \times\frac{3\:mol\:CO_{2}}{1\:mol\:C_{3}H_{8}}\times\frac{44g\:CO_{2}}{1 mol\:CO_{2}} = 300 g\:CO_{2}](https://tex.z-dn.net/?f=100%20g%5C%3Aof%5C%3AC_%7B3%7DH_%7B8%7D%20%5Ctimes%5Cfrac%7B1%5C%3Amol%5C%3AC_%7B3%7DH_%7B8%7D%7D%7B44%20g%5C%3AC_%7B3%7DH_%7B8%7D%7D%20%5Ctimes%5Cfrac%7B3%5C%3Amol%5C%3ACO_%7B2%7D%7D%7B1%5C%3Amol%5C%3AC_%7B3%7DH_%7B8%7D%7D%5Ctimes%5Cfrac%7B44g%5C%3ACO_%7B2%7D%7D%7B1%20mol%5C%3ACO_%7B2%7D%7D%20%3D%20300%20g%5C%3ACO_%7B2%7D)
The equation of the reaction of copper and silver nitrate is given as follows:
![Cu + 2\:AgNO_{3}\rightarrow Cu(NO_{3})_{2} + 2\:Ag](https://tex.z-dn.net/?f=Cu%20%2B%20%202%5C%3AAgNO_%7B3%7D%5Crightarrow%20Cu%28NO_%7B3%7D%29_%7B2%7D%20%2B%202%5C%3AAg)
![5.7 g\:Cu \times\frac{1\:mol\:Cu}{64 g\:Cu} \times\frac{1mol\:Cu(NO_{3})_{2}}{1mol\:Cu}\times\frac{188g\:Cu(NO_{3})_{2}}{1 mol\:Cu(NO_{3})_{2}} = 16.7 g\:Cu(NO_{3})_{2}](https://tex.z-dn.net/?f=5.7%20g%5C%3ACu%20%5Ctimes%5Cfrac%7B1%5C%3Amol%5C%3ACu%7D%7B64%20g%5C%3ACu%7D%20%5Ctimes%5Cfrac%7B1mol%5C%3ACu%28NO_%7B3%7D%29_%7B2%7D%7D%7B1mol%5C%3ACu%7D%5Ctimes%5Cfrac%7B188g%5C%3ACu%28NO_%7B3%7D%29_%7B2%7D%7D%7B1%20mol%5C%3ACu%28NO_%7B3%7D%29_%7B2%7D%7D%20%3D%2016.7%20g%5C%3ACu%28NO_%7B3%7D%29_%7B2%7D)
Therefore, the mass of oxygen required is 17.06 g; the mass of carbon dioxide produced is 300 g; the mass of copper (ii) nitrate produced is 16.7 g.
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1.09 Liters
The total volume of the solution formed is 1.093 L
We know that,
Ca(OH)2 = 74 g/mole
n = 55/74 = 0.743 mole
M = n/V
where,
M = mass of the solution
n = number of moles
V = Volume of solution
0.680 = 0.743 / V
V = 0.743/0.680
= 1.093 L
= 1093 ml
Therefore, the total volume of the solution formed is 1.093 L
<h3>What is molarity?</h3>
- The amount of a substance in a specific volume of solution is known as its molarity (M).
- The number of moles of a solute per liter of a solution is known as molarity.
- The molar concentration of a solution is another name for molarity.
<h3>What is the molarity unit?</h3>
- Molarity is measured in mol/L.
- The molarity, which is given as moles/liter, is the quantity of moles of solute per liter of the solution.
<h3>Example of molarity:</h3>
- The moles of solute per liter of solution is measured as molarity.
- For instance, when table salt is dissolved in water, water serves as both the solution and the solute.
- Sodium chloride weights 58.44 grams per mole.
- One molar solution, or 1M, is created by dissolving 58.44 grams of sodium chloride in one liter of water.
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