What element appears before iron

A 4.7-L sealed bottle containing 0.33 g of liquid ethanol,C₂H₆O, is placed in a refrigerator and reaches equilibrium with its va

Mama L [17]

The mass of ethanol present in the vapor is 8.8×10⁻²g. when liquid and vapor ethanol at equilibrium.

The volume of the bottle = 4.7 L

Mass of ethanol = 0.33 g

Temperature (T1) = -11 oC = 273-11 = 262 K

P1 = 6.65 torr

Now we will calculate the mole by applying the ideal gas equation:-

PV = nRT

Or, n = PV/RT

Where P is the pressure

T is the temperature

R is the gas constant = 0.0821 L atm mol-1K-1

V is the volume

Substituting the values of P, V, T, and R the mole of ethanol is calculated as:-

= 0.001913 mol C2H6

Conversion of the mole to gm

Molar mass of ethanol (M) = 46.07 g/mol

Mass of C2H6O =0.001913 mol C2H6O 46.07 g/mol = 0.088 = 8.8×10⁻²g.

Hence, the mass of ethanol present in the vapor is found to be 8.8×10⁻²g.

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8 0

2 years ago

What is the maximum number of d orbitals in a principal energy level?

Alenkasestr [34]

The maximum number of D orbitals in a principal energy level is 5.

7 0

3 years ago

Read 2 more answers

What is the speed of a horse in meters per second that runs a distance of 1.2 miles in 2.4 minutes? (Hint: 1 mile = 1609 m)

dezoksy [38]

Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]

7 0

4 years ago

If boron atoms have 5 electrons, how many electrons are in the outer electron shell of a boron atom?

ANEK [815]

Answer:

3

Explanation:

The first election shell can only hold 2 electrons, but the next one can hold up to 8

3 0

2 years ago

Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th

ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol$

The chemical equation for the reaction of ammonia and HCl follows:

$NH_3+HCl\rightarrow NH_4^++Cl^-$

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonia

Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

$\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M$

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = $\frac{1}{1}\times 0.0035=0.0035mol$ of ammonium ion

Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

$\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.8\times 10^{-5}$

$10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}$

The chemical equation for the dissociation of ammonium ion follows:

$NH_4^+\rightarrow NH_3+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[NH_3][H^+]}{[NH_4^+]}$

We know that:

$[NH_3]=[H^+]=x$

$[NH_4^+]=0.072M$

Putting values in above expression, we get:

$5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M$

To calculate the pH concentration, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=6.32\times 10^{--6}M$

$pH=-\log (6.32\times 10^{-6})\\\\pH=5.20$

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0

3 years ago