How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

Molarity is defined as the number of moles present in one liter of solution. Given the mass of NH₃ is 2.35 g and its molar mass is 17 g/mol, the moles of NH₃ in 2.35 g can be calculated as 2.35 g / 17 g/mol = 0.138 mol. Consequently, in a 0.05 L solution, the number of moles amounts to 0.138 mol. Therefore, the concentration in 1 L is: 0.138 mol / 0.05 L x 1L = 2.76 mol. Thus, the molarity of NH₃ is 2.76 M.

Answer:

The volume of calcium hydroxide solution utilized is 0.0235 mL.

Explanation:

Moles of KHP =

In accordance with the reaction, 2 moles of KHP react with 1 mole of calcium hydroxide, thus 0.0330 moles of KHP will react with;

of calcium hydroxide

The molarity of calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

The volume of the calcium hydroxide solution utilized is 0.0235 mL.

Answer:

HCl and CaCl2

Explanation:

Calcium hydroxide acts as a base, thus requiring an acid for neutralization.

HCl functions as an acid, which can neutralize calcium hydroxide when they react together.

The resulting salt will be calcium chloride (CaCl2).

The interaction between calcium hydroxide and HCl yields Ca(OH)Cl and CaCl2.

Here, Ca(OH)Cl exists as a solution and calcium chloride serves as a base.

I hope this response assists you.

Answer:

22.7

Explanation:

Initially, calculate the energy released from the sample's mass. The combustion heat represents the energy per mole of the fuel:

ΔHC=qrxnn

We can rearrange this formula to isolate qrxn, remembering to convert the sample's mass into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released during the reaction must match the total heat absorbed by both the water and the bomb calorimeter:

qrxn=−(qwater+qbomb)

The heat absorption by the water can be calculated using its specific heat:

qwater=mcΔT

The calorimeter's heat absorption can be derived from its heat capacity:

qbomb=CΔT

Combine both equations into the first equation, substituting the known details, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute each multiplication term and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)−(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)−(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)−50208 J+(420. J∘C×Tfinal)−8400 J]

Combine the similar terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, isolate Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

Round to three significant figures gives the final result as 22.7∘C.