How many atoms are present in the following compound?<br> C21H22N2O2

Calculate the pH of each solution at 25∘C

JulijaS [17]

Answer: pH of HCl =5, HNO3 = 1,

NaOH = 9, KOH = 12

Explanation:

pH = -log [H+ ]

1. 1.0 x 10^-5 M HCl

pH = - log (1.0 x 10^-5)

= 5 - log 1 = 5

2. 0.1 M HNO3

pH = - log (1.0 x 10 ^ -1)

pH = 1 - log 1 = 1

3. 1.0 x 10^-5 NaOH

pOH = - log (1.0 x 10^-5)

pOH = 5 - log 1 = 5

pH + pOH = 14

Therefore , pH = 14 - 5 = 9

4. 0.01 M KOH

pOH = - log ( 1.0 x 10^ -2)

= 2 - log 1 = 2

pH + pOH = 14

Therefore, pH = 14 - 2 = 12

8 0

3 years ago

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol$

To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0

3 years ago

Use the equation: 2 kclo3 2 kcl + 3 o2 to write the mole ratio needed to calculate: (a) moles of kcl produced from 3.5 moles of

poizon [28]

Answer:-

(a) 3.5

(b) 3

Explanation:-

2KClO3 --> 2KCl + 3 O2

From the equation we see that 2 moles of KClO3 gives 2 moles of KCl.

So 3.5 moles of KClO3 will give 3 moles of KCl.

Again

3 moles of O2 are produced with 2 moles of KCl.

If 4.5 moles of O2 produced then

moles of KCl = 4.5 x 2/3

=3

3 0

3 years ago

What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?

Nataly_w [17]

The glass opposite to the negative electrode started to glow. Hence, option B is correct.

<h3>What is a cathode ray tube?</h3>

A cathode-ray tube (CRT) is a specialized vacuum tube in which images are produced when an electron beam strikes a phosphorescent surface.

J.J. Thomson, through his famous Cathode ray experiment, proved that all atoms contain small negatively charged particles known as electrons. In the experiment, he applied electric voltage across a cathode ray tube. a fluorescent material coating was done on the positive side. When the voltage was applied, the positive side has glowing dots.

Hence, option B is correct.

Learn more about the cathode ray tube here:

brainly.com/question/14409449

#SPJ1

6 0

2 years ago

Trying to take the ged test please help me I’m stuck

JulsSmile [24]

Answer: C) Either benzene or oxygen may limit the amount of product that can be formed

Explanation: Benzene and oxygen are the reactants of the equation. What type and the amount of reactants there are in a chemical reaction affects the outcome. Therefore, seeing as benzene and oxygen are the reactants, the answer is C).

3 0

3 years ago

How many atoms are present in the following compound? C21H22N2O2