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3 years ago

Calculate the number of moles of Li3PO4 in 2.2 L of a 0.60 M Li3PO4 solution.

Chemistry

1 answer:

Bingel [31]3 years ago

4 0

Answer:

1.32 mole

Explanation:

The following data were obtained from the question:

Volume of solution = 2.2L

Molarity of solution = 0.60M

Mole of Li3PO4 =..?

Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is represented as:

Molarity = mole /Volume

With the above formula we can easily calculate the number of mole of Li3PO4 as shown below:

Molarity =mole /Volume

0.6 = mole of Li3PO4 /2.2

Cross multiply

Mole of Li3PO4 = 0.6 x 2.2

Mole of Li3PO4 = 1.32 mole

Therefore, 1.32 mole of Li3PO4 is contained in the solution.

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When heated, allyl aryl ethers and allyl vinyl ethers undergo a reaction called a Claisen rearrangement, a concerted reorganizat

Alex_Xolod [135]

Answer:

See explanation below

Explanation:

I found a picture of this exercise, to show you how to do this problem.

The first picture is the compound, and the second is the mechanism of reaction to do this claisen rearrangement.

Hope this helps

5 0

3 years ago

The pH of a 0.65M solution of hydrofluoric acid HF is measured to be 1.68. Calculate the acid dissociation constant Ka of hydrof

sergey [27]

Answer:

Kₐ = 6.7 x 10⁻⁴

Explanation:

First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:

HF + H₂O ⇄ H₃O⁺ + F⁻

Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]

Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.

pH = -log [ H₃O⁺ ]

1.68 = - log [ H₃O⁺ ]

Taking antilog to both sides of this equation:

10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]

[ F⁻ ] = 2.1 X 10⁻² M

Solving for Kₐ :

Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴

(Rounded to two significant figures, the powers of 10 have infinite precision )

4 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

Leya [2.2K]

Answer:

See explanation

Explanation:

The molecular equation shows all the compounds involved in the reaction.

The molecular equation is as follows;

2NaF(aq) + Pb(NO3)2(aq) -------> PbF2(s) + 2NaNO3(aq)

The complete ionic equation shows all the ions involved in the reaction

The complete ionic equation;

2Na^+(aq) + 2F^-(aq) + Pb^2+(aq) + 2NO3^-(aq) -------->PbF(s) + 2Na^+(aq) +2NO3^-(aq)

The net Ionic equation shows the ions that actually participated in the reaction

The net ionic equation is;

2F^-(aq) + Pb^2+(aq)--------> PbF(s)

3 0

3 years ago

Element c has a strong emmission aound 450nm Does this emmission line reprosent a lower of higher emergy line at 627 nm?

Julli [10]

It respresents the higher energy level than 627nm .

<h3>What is a emission line ? </h3>

Emission lines are the glowing hot gas emits lines of light whereas absorption line refers to the tendency of cool atmospheric gas to absorb the same line of light.Some lights produce dark band when the light passes through gas in the atmosphere . There are two line spectrum and absorption.

spectrum is an excitement of electrons from lower to higher energy levels and when it comes back it releases energy in the terms of colourful lights .

It represents the higher energy levels than 627nm because Energy is inversly proportional to wavelength .

to learn more about Emission lines click here

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5 0

2 years ago