0.444 mol C2H5OH= how many molecules

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0

2 years ago

Please help!! Balancing Nuclear Equations

sineoko [7]

The missing part of the equation is found to be 4/2He. Option A

<h3>What are nuclear equations?</h3>

The term nuclear equations have to do with the type of equation in which one type of nucleus is transformed into another sometimes by the bombardment or loss of a particle.

Now the full equation ought to be written as 7/3Li + 1/1H -----> 4/2He + 4/2He. This is because the total mass on the left is 8 and the total charge on the left is 4.

Learn more about nuclear equations:brainly.com/question/19752321

#SPJ1

8 0

1 year ago

List 2 extensive properties and 3 intensive properties of a 5.0g, 1 cm3 cube of silver

jeka57 [31]

Properties of matter can be broadly classified into two categories:

Physical properties which usually involve a change in the state of matter and Chemical properties which involve a change in the chemical composition of matter.

Now, physical properties can be further classified as:

Extensive: these depend on the amount of the substance, eg: mass, volume

Intensive: these do not depend on the amount of the substance eg: density, color, melting point, boiling point

Here we are given a 5.0 g and 1 cm3 silver cube :

Therefore:

Extensive properties are-

1) Mass of silver = 5.0 g

2) Volume of silver = 1 cm3

Intensive properties are:

1) Density of silver = mass/volume = 5.0 g/ 1 cm3 = 5.0 g/cm3

2) Melting point of silver = 962 C

3) Color = white/gray

3 0

2 years ago

Write a balanced overall reaction from these unbalanced half-reactions sn and ag

loris [4]

Balance each one by adding electrons to make the charges on both sides the same:

Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag

Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:

2 Ag+ + 2 e- --> 2 Ag

Now, just add the two half reactions together, cancelling anything that's the same on both sides:

2 Ag+ + Sn --> Sn2+ + 2 Ag

And you're done.

8 0

2 years ago

If an acid/base dissociates completely in water what type of substance is it? Does this mean the acid/base is dangerous?

hram777 [196]

Answer:

1 strong acid

2 yes they are dangerous

Explanation:

Since nearly all of it is dissociated in water, it is called a strong acid.

2 yes Concentrated strong acids can cause severe and painful burns. The pain is due in part to the formation of a protein layer, which resists further penetration of the acid

3 0

2 years ago