Enyzmes change the _______ of a reaction by making the reaction go _____

Hydrogen gas and fluorine gas will react to form hydrogen fluoride gas. What is the standard free energy change for this reactio

Marta_Voda [28]

Answer:

$\Delta G=-541.4kJ/mol$

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

$H_2+F_2\rightarrow 2HF$

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

$\Delta G=2\Delta G_f^{HF}-(\Delta G_f^{H_2}+\Delta G_f^{F_2})\\\\\Delta G=2*-270.70kJ/mol-(0kJ/mol+0kJ/mol)\\\\\Delta G=-541.4kJ/mol$

Regards!

4 0

3 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

6 0

3 years ago

3. If we remember seeing something happen, we can trust that it happened just as we think it did. True or False

djverab [1.8K]

$\huge{ \underbrace{ \overbrace{True}}}$

.

$\huge\mathfrak\purple{Hope \: it \: helps}$

Note : Because Many of My Dreams came true and I realized.

3 0

3 years ago

What molecule experience hydrogen bonding

enot [183]

Hydrogen bonds to either Nitrogen, Oxygen, or Fluorine to experience Hydrogen bonding.

3 0

3 years ago

A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an

inysia [295]

<u>Answer:</u> The molar mass of the unknown compound is 223.2 g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 8.44 torr

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $62.3637\text{ L torr }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = $4.54\times 10^{-4}M$

Given mass of unknown compound = 15.2 mg = 0.0152 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 150.0 mL

Putting values in above equation, we get:

$4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol$

Hence, the molar mass of the unknown compound is 223.2 g/mol

5 0

3 years ago

Enyzmes change the _______ of a reaction by making the reaction go ______

Enyzmes change the _ of a reaction by making the reaction go