How many moles of PCl3 are produced if 5 moles of P4 reacted with 22 moles of Cl2

A solution is made by mixing equal masses of methanol, CH4O, and ethanol, C2H6O. Determine the mole fraction of each component t

Serggg [28]

Answer: mole fraction of methanol = 0.590

mole fraction of ethanol = 0.410

Explanation:

We are given:

Equal masses of methanol $CH_4O$ and ethanol $C_2H_6O$ are mixed.

let the mass be x g.

Calculating the moles of methanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{32.04g/mol}$

Calculating the moles of ethanol in the solution, by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{xg}{46.07g/mol}$

To calculate the mole fraction of methanol, we use the equation:

$\chi_{methanol}=\frac{n_{methanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{methanol}=\frac{\frac{xg}{32.04g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.590$

To calculate the mole fraction of ethanol, we use the equation:

$\chi_{ethanol}=\frac{n_{ethanol}}{n_{methaol}+n_{ethanol}}$

$\chi_{ethanol}=\frac{\frac{xg}{46.07g/mol}}{\frac{xg}{32.04g/mol}+\frac{xg}{46.07g/mol}}=0.410$

Thus mole fraction of methanol is 0.590 and mole fraction of ethanol 0.410 in three significant figures.

5 0

2 years ago

1) Sodium hydroxide is deliquescent. A sample of 3.0 g was dissolved in 100 mL; 10 mL was titrated with 34.9 mL HCl 0.2 M. What

ELEN [110]

From the calculation, the percentage of water in the sodium hydroxide sample is 7%.

<h3>What is neutralization?</h3>

The term neutralization has to do with the reaction between an acid and a base to yiled salt and water.

Now we have to apply the titration formula;

CAVA/CBVB = NA/NB

CA = concentration of acid

CB = concentration of base

VA = volume of acid

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

The reaction equation is; HCl + NaOH ----->NaCl + H2O

CAVANB = CBVBNA

CB = CAVANB /VBNA

CB = 34.9 * 0.2 M * 1/10 * 1

CB = 0.698 M

Number of moles = Conncentration * volume

= 0.698 M * 100/1000 L = 0.0698 moles

Mass = Number of moles * molar mass

Mass = 0.0698 moles * 40 g/mol = 2.79 g

percent of NaOH = 2.79 g/ 3g * 100/1 = 93%

Percent of water = 100- 93 = 7%

Learn more about neutralization: brainly.com/question/15395418

5 0

1 year ago

What is a proper adjective?

Alex17521 [72]

This is what a proper adjective is.

7 0

2 years ago

For each solution, determine the p-values for each ion indicated. A solution that is 0.493 M in NaCl and 0.314 M in NH 4 Cl .

miss Akunina [59]

Complete Question:

Ions to calculate the p-values: Na⁺, Cl⁻, and NH₄⁺

Answer:

pNa = 0.307

pCl = 0.093

pNH₄ = 0.503

Explanation:

The p-value is calculated by the antilog of the concentration of the substance of interest. For example, pH = -log[H⁺]. Thus, first, let's find the ions concentration.

Both substances are salts that solubilize completely, thus, by the solution reactions:

NaCl → Na⁺ + Cl⁻

NH₄Cl → NH₄⁺ + Cl⁻

So, for both reactions the stoichiometry is 1:1:1 and the concentration of the ions is equal to the concentration of the salts.

[Na⁺] = 0.493 M

[Cl⁻] = 0.493 + 0.314 = 0.807 M

[NH₄⁺] = 0.314 M

The p-values are:

pNa = -log[Na⁺] = -log(0.493) = 0.307

pCl = -log[Cl⁻] = -log(0.807) = 0.093

pNH₄ = -log[NH₄⁺] = -log(0.314) = 0.503

7 0

3 years ago

A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.

Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

5 0

3 years ago