20g of ideal gas contains only atoms of s and o occupies 5.6l at ntp what is the mol wt of gas

In a car piston shown above, the pressure of the compressed gas (red) is 5.00 atm. If the area of the piston is 0.0760 m^2. What

Anarel [2989]

Answer:

The force is 38503.5N.

Explanation:

From the problem, we determine:

P (pressure) = 5.00 atm.

Next, to find the force in Newtons (N), we must convert 5 atm into N/m², as shown:

1 atm equals 101325 N/m².

So, 5 atm equals 5 x 101325 = 506625 N/m².

A (the piston area) = 0.0760 m².

Pressure signifies force per unit area, mathematically represented as

P = F/A.

From this, we find F = P × A.

F = 506625 × 0.0760.

Therefore, F = 38503.5N.

Thus, the piston experiences a force of 38503.5N.

6 0

3 months ago

The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan

eduard [2782]

Response: The rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

Rationale:

Based on the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant when $701K$ = $2.57M^{-1}s^{-1}$

$K_2$ = rate constant when $525K$ =?

$Ea$ = activation energy for the process = $1.5\times 10^2kJ/mol=1.5\times 10^5J/mol$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 701 K

$T_2$ = final temperature = 525 K

Substituting the provided values into this formula yields:

$\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]$

$K_2=0.0606M^{-1}s^{-1}$

Thus, the rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

8 0

2 months ago

In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OAT

lorasvet [2795]

Answer: Reaction 2 is a spontaneous one.

Explanation:

According to our understanding:

$\Delta G$ = +ve, meaning the reaction is non-spontaneous

$\Delta G$ = -ve, indicating the reaction is spontaneous

$\Delta G$ = 0, stating that the reaction is at equilibrium

For a reaction to be classified as spontaneous, the Gibbs free energy must yield a negative value.

Reaction 1:

Glucose + Pi ⟶ glucose-6-phosphate + H₂O, ΔG = +13.8 kJ/mol $\rightarrow$

Reaction 2:

ATP + H₂O ⟶ ADP + Pi, ΔG = -30.5 kJ/mol $\rightarrow$

From this, we can conclude that ΔG being negative indicates that reaction 2 is indeed spontaneous.

8 0

1 month ago