CHEMISTRY COLLEGE

Consider the reaction at 25 °C. H2O(l) ↔ H2O(g) ΔG° = 8.6 kJ/mol Calculate the pressure of water at 25 °C (Hint: Get K eq)

Answers

Answer 1
Answer:

Answer:

\boxed{\text{23.4 mmHg}}

Explanation:

H₂O(ℓ) ⟶ H₂O(g)

K_{\text{p}} = p_{\text{H2O}}

\text{The relationship between $\Delta G^(\circ)$ and $K_{\text{ p}}$ is}\n\Delta G^(\circ) = -RT \ln K_{\text{p}}

Data:  

T = 25 °C

ΔG° = 8.6 kJ·mol⁻¹

Calculations:

T = (25 + 273.15) K = 298.15 K

\begin{array}{rcl}8600 & = & -8.314 * 298.15 \ln K \n8600 & = & -2478.8 \ln K\n-3.47 & = & \ln K\nK&=&e^(-3.47)\n& = & 0.0311\end{array}

Standard pressure is 1 bar.

p_{\text{H2O}} = \text{0.0311 bar} * \frac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\n\n\text{The vapour pressure of water at $25 ^(\circ)\text{C}$ is $\boxed{\textbf{23.4 mmHg}}$}

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A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?​

Answers

Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

What is the symbol of the element that is classified as an alkali metal and is in period 4?

Answers

Answer:

Potassium (K) [First element in period 4]

Which step would help a student find the molecular formula of a compound from the empirical formula? Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Subtract the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound from the subscripts of the empirical formula. Divide the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Add the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound to the subscripts of the empirical formula..

Answers

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

got it right on edge 2020 :)

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

Which of the following shows the abbreviation for an SI unit of density?

Answers

Is there supposed to be a graph or something?

the aswer is g/ml hope is helpful

Aqueous solutions of sodium hypoch lorite (NaOCI), best known as bleach, are prepared by the reaction of sodium hydroxide with chlorine: 2 NaOH (aq)Cl2(g)->NaOCI (aq)+ H20 (I)+ NaCl (aq) How many grams of NaOH are needed to react with 25.0 g of chlorine?

Answers

Answer:

28.2 g of NaOH

Explanation:

We need to calculate the grams of NaOH needed to react with 25.0 g of Cl₂ in the following reaction:

2 NaOH(aq) + Cl₂(g) → NaOCI(aq + H₂0(I) + NaCl(aq)

We are going to solve this by making use of the molar ratio between Cl₂ and NaOH given by the reaction equation where we see that every mol of Cl₂ will react with 2 moles of NaOH.

So first we need to convert the 25.0 g of Cl₂ to moles:

  • Molar Mass of Cl₂ = 2 x 35.45 = 70.90 g/mol
  • Moles of Cl₂ = 25.0 g / 70.90 g/mol = 0.3526 moles

Then we need to calculate the moles of NaOH needed to react with these moles of Cl₂ knowing that every mol of Cl₂ will react with 2 moles of NaOH:

  • moles of NaOH = 2 x moles of Cl₂ = 2 x 0.3526 moles = 0.7052 moles

Next we must convert these moles to grams:

  • Molar Mass of NaOH = 22.990 + 15.999 + 1.008 = 40.00 g/mol
  • Mass of NaOH = 0.7052 moles x 40.00 g/mol = 28.2 g

28.2 g are needed to react with 25.0 g of Cl₂ in the production of NaOCl

If the heat of combustion of carbon monoxide (CO) is −283.0 kJmole, how many grams of carbon monoxide must combust in order to release 2,500.kJ of energy?

Answers

Answer:

247.4 g

Explanation:

Let's consider the thermochemical equation for the combustion of carbon monoxide.

CO(g) + 0.5 O₂(g) ⇒ CO₂(g)    ΔH°c = -283.0 kJ/mol

The moles of carbon monoxide required to release 2500 kJ (-2500 kJ) are:

-2500 kJ × (1 mol CO/-283.0 kJ) = 8.834 mol CO

The molar mass of CO is 28.01 g/mol. The mass corresponding to 8.834 moles of CO is:

8.834 mol × 28.01 g/mol = 247.4 g
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