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2 years ago

By titration, it is found that 28.5 mL of 0.183 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the

Chemistry

1 answer:

maksim [4K]2 years ago

4 0

Answer:

0.209M

Explanation:

M1V1=M2V2

(28.5 mL)(0.183M)=(25.0mL)(M)

M2= 0.209M

*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)

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Consider the chemical reaction represented below

hjlf

Answer:

Explanation:

4 0

3 years ago

I need help with this question

Maru [420]

Aluminum sulfide : Al₂S₃

ratio cation : anion = 2 : 3

<h3>Further explanation</h3>

Given

Compound of Aluminium

Required

cations anions ratio

Solution

Salt can be formed from cations and anions which have their respective charges.

In the chemical compound formula these charges are crossed with each other

For aluminum it has a +3 charge

1. Aluminum carbide : Al₄C₃

ratio cation : anion = 4 : 3

2. Aluminum chloride : AlCl₃

ratio cation : anion = 1 : 3

3. Aluminum sulfide : Al₂S₃

ratio cation : anion = 2 : 3

4. Aluminum nitride : AlN

ratio cation : anion = 1 : 1

3 0

2 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

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

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

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

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

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

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago

What is the correct balanced equation for the following equilibrium constant expression?

weeeeeb [17]

The correct answer is 3) 2CO2(g) ⇄ 2CO(g) + O2(g)

this is the correct one because it is a decomposition reaction and all the number of atoms is equal on both sides.

there are 2 C atoms on both sides.

and 4 O atoms on both sides.

and 1) the atoms numbers are equal on both sides but not correct as it not a

correct number as it has 1/2 O2.

and 2) CO2(g) ⇆ CO(g) + O2

the number of O atoms is not equal on both sides of the equation.

we have 2 O atoms on the left side and 3 O atoms on the right side.

so, this not a balanced equation.

4) also not correct 2CO(g) + O2 ⇆ 2CO2

as it is not a decomposition reaction and the 2CO & O2 are as reactants not products.

so the correct answer is 3) 2CO2(g) ⇆ 2CO(g) + O2(g)

8 0

3 years ago

• Draw lewis dot structures for an atom of each of following elements:

marin [14]

Answer:

Explanation:

Lewis dot structures represent the symbol of an atom we're looking at and the number of valence electrons it has. This number is represented by the sum of dots around the symbol.

Potassium is in group 1A, this means it only has one valence electron, so we draw K with one dot in its Lewis diagram;
Argon is in group 8A, this means it has eight valence electrons, so we draw Ar with 8 dots around it in its Lewis diagram;
Silicon is in group 4A, this means it has four valence electrons, so we draw Si with 4 dots around it in its Lewis diagram;
Arsenic is in group 5A, this means it has five valence electrons, so we draw As with 5 dots around it in its Lewis diagram.

Those are represented in the image attached below:

5 0

2 years ago