The normal boiling point of c2cl3f3 is 47.6°c and its molar enthalpy of vaporization is 27.49 kj/mol. what is the change in entr

The response is:

No, the equation is not balanced. Neither the Nitrogen (N) nor the Hydrogen (H) are in balance!

Here's the reasoning:

⓵ A properly balanced chemical equation means that the quantity of atoms on the reactants side matches that on the products side.

→ The equation lacks balance because there are 2 Nitrogen atoms and 2 Hydrogen atoms on the reactants side. In contrast, on the products side, there is only 1 Nitrogen atom and 4 Hydrogen atoms. Thus, the number of atoms on each side is not consistent!

Hopefully, this clarification is helpful; feel free to reach out if you have any further questions! ☻

I hope this is useful......

Answer: C. 151 g

Solution: The balanced equation given is:

From this equation, the ratio of moles between all substances is 1:1. We have 117 grams of LIOH and 141 grams of HBr available and need to calculate the theoretical yield of LiBr.

We should convert each reactant’s grams into moles to identify the limiting reagent since the theoretical yield relies on it.

Molar mass for LiOH = 6.94 + 15.999 + 1.008 = 23.947 grams per mole

Molar mass for HBr = 1.008 + 79.904 = 80.912 grams per mole

To find the moles of each reactant, we divide their grams by their respective molar masses.

Moles of LiOH = = 4.89 mol

Moles of HBr = = 1.74 mol

As there are fewer moles of HBr, it is the limiting reactant. With a 1:1 mol ratio between HBr and LiBr, 1.74 moles of LiBr can be produced.

Molar mass of LiBr = 6.94 + 79.904 = 86.844 grams per mole

The mass of LiBr formed = = 151 g LiBr

Based on calculations, the theoretical yield of LiBr is 151 g, hence the correct answer is C.

Answer:

The solution composed of 35.0 g of mixed with 250.0 g of ethanol will exhibit the lowest freezing temperature

Explanation:

where,

= reduction in freezing point =  

= freezing point constant  

m = molality

As observed, a higher molality in the solution corresponds to a greater depression in the freezing point, resulting in a more reduced freezing point of the solution.

A. 35.0 g of in 250.0 g of ethanol.

The amount of moles of =

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

B. 35.0 g of in 250.0 g of ethanol

Moles of

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

C. 35.0 g of in 250.0 g of ethanol

Moles of =

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

The solution containing 35.0 g of added to 250.0 g of ethanol will yield the lowest freezing point