What is the area beyond the black hole where the black hole has stopped expanding and matter will no longer be pulled in?

A certain reaction at equilibrium has more moles of gaseous products than of gaseous reactants.(b) Write a statement about the r

mojhsa [17]

To be equal as $K_p$ , $K_c$ needs to be divided by $(RT)^{\Delta n}$

<h3>Briefly explained</h3>

When the amount of gaseous reactant is greater than the amount of gaseous product, where $n_{products} < n_{reactants}$ ,

then = $K_p < K_c$

It's because $\Delta n$ is negative, which places (RT) in the denominator. This is how the equation will now appear.

$K_p = K_c(RT)^{-\Delta n} = \frac{ K_c}{(RT)^{\Delta n}}$

Here you can observe the value of $K_c$ is greater than $K_p$ . To be equal as $K_p$ , $K_c$ needs to be divided by $(RT)^{\Delta n}$

<h3>What is Δngas?</h3>

The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.

Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.

<h3>Definition of equilibrium</h3>

When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.

Learn more about equilibrium

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6 0

2 years ago

Calculate the mass of water produced when 9.57 g of butane reacts with excess oxygen.

irinina [24]

Answer:

14.9 g

Explanation:

Step 1: Write the balanced equation

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

9.57 g × 1 mol/58.12 g = 0.165 mol

Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀

0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O

Step 4: Calculate the mass corresponding to 0.825 mol of H₂O

The molar mass of H₂O is 18.02 g/mol.

0.825 mol × 18.02 g/mol = 14.9 g

4 0

3 years ago

What would be the final temperature of a 73.174g sample

madreJ [45]

The final temperature is -138 °C.

Explanation:

Using the equation of specific heat

$Q = m*c* del T$

We can easily find the final temperature of a 73.174 g of copper sample. As we know that specific heat is the amount of energy required to raise the temperature of the object to 1°C.

The specific heat of copper is known as 0.387 J/g°C and the initial temperature is said as 102 °C . The mass is given as 73.174 g. The heat released is 6800 J.

Since the heat is released the Q value will be negative.

$-6800=0.387*73.174*(T_{final}-102)$