WILL GIVE BRAINLIEST

For the following reaction, KP = 0.455 at 945°C: At equilibrium, is 1.78 atm. What is the equilibrium partial pressure of CH4 in

Alborosie

Answer:

See explanation below

Explanation:

The question is incomplete. However, here's the missing part of the question:

"For the following reaction, Kp = 0.455 at 945 °C:

C(s) + 2H2(g) <--> CH4(g).

At equilibrium the partial pressure of H2 is 1.78 atm. What is the equilibrium partial pressure of CH4(g)?"

With these question, and knowing the value of equilibrium of this reaction we can calculate the partial pressure of CH4.

The expression of Kp for this reaction is:

Kp = PpCH4 / (PpH2)²

We know the value of Kp and pressure of hydrogen, so, let's solve for CH4:

PpCH4 = Kp * PpH2²

*: You should note that we don't use Carbon here, because it's solid, and solids and liquids do not contribute in the expression of equilibrium, mainly because their concentration is constant and near to 1.

Now solving for PpCH4:

PpCH4 = 0.455 * (1.78)²

PpCH4 = 1.44 atm

6 0

3 years ago

N2 + 3H2 = 2NH3

My name is Ann [436]

This is so basic bro u gotta first move

6 0

3 years ago

Neutralizing an olympic size swimming pool is conceptually very similar to performaing a massive titration experiment. Suppose a

MrRissso [65]

Answer:

6,97x10⁻³ gallons

Explanation:

pH is defined as:

pH = -log [H⁺]

Thus, you need to have, in the end:

10⁻⁷ = [H⁺]

And you have, in the first:

$10^{-9,33}$ = [H⁺]

The volume of swimming pool is:

700'000 galllons × $\frac{3,78541 L}{1 gallon}$ = 2649787 L

Thus, the moles of H⁺ in the first and in the end are:

First:

$10^{-9,33}mol/L$ × 2649787L = 1,24x10⁻³ moles

End:

$10^{-7}mol/L$ × 2649787L = 0,265 moles

Thus, the moles of H⁺ you need to add are:

0,265 - 1,24x10⁻³ = 0,26376 moles

These moles comes from 10M HCl, thus, the volume in gallons you need to add are:

$0,26376moles*\frac{1L}{10moles}* \frac{1gallon}{3,78541L}$ =

6,97x10⁻³ gallons

I hope it helps!

7 0

3 years ago

The energy source used in nuclear power plants today is

Klio2033 [76]

Answer:

B I have taken the quiz already 90%

7 0

3 years ago

If 42.7 of 0.208 M hydrochloric acid are needed to completely neutralize a solution of calcium hydroxide, how many grams of calc

Montano1993 [528]

Answer:

0.329 g

Explanation:

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:

$Ca(OH)_2(aq)+2 HCl (aq)\rightarrow CaCl_2 (aq)+2 H_2O (l)$

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

$c_{HCl}=\frac{n_{HCl}}{V_{HCl}}$