What mass of fluorine is needed to completely react with 3.6 grams of water?

200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentratio

Ede4ka [16]

Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

First we calculate how many moles of each reagent were added:

HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl

Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we calculate how many moles of Ba(OH)₂ reacted, by converting the total number of HCl moles to Ba(OH)₂ moles:

203.85 mmol HCl * $\frac{1mmolBa(OH)_{2}}{2mmolHCl}$ = 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

There are two OH⁻ moles per Ba(OH)₂ mol:

OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

Finally we divide the number of OH⁻ moles by the total volume (100 mL + 200 mL):

716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

7 0

3 years ago

To balance a chemical equation, it may be necessary to adjust the

lana66690 [7]

Answer:

A. Coefficients

Explanation:

that's the number in front of the molecules

7 0

3 years ago

Which of the following is a type of cell organelle?

icang [17]

Answer:

C . Dna

Explanation:

8 0

3 years ago

PLEASE HELP ASAP ESSAY WORTH 15 POINTS! ONLY GENIUS!

Sliva [168]

Answer:

According to the law of conservation of mass, the mass of reactants will be equal to the mass of the products.

Explanation:

5 0

3 years ago

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding

slega [8]

Answer:

$Binding\ energy=43.43\times 10^{-20}\ J$

Explanation:

Using the expression for the photoelectric effect as:

$E=h\nu_0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

$\nu_0=\frac {c}{\lambda_0}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light being bombarded

Given, $\lambda=4.00\times 10^{-7}\ m$

$\frac {hc}{\lambda_0}$ is the binding energy or threshold energy

$\frac {1}{2}\times m\times v^2$ is the kinetic energy of the electron emitted. = $6.26\times 10^{-20}\ J$

Thus, applying values as:

$\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy$

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$Binding\ energy=43.43\times 10^{-20}\ J$

5 0

3 years ago