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Fynjy0 [20]
3 years ago
10

If the gram-formula mass of substance X is 180 g/mol, determine the molarity of the solution at Point E

Chemistry
1 answer:
Alina [70]3 years ago
7 0
<h3><u>Answer;</u></h3>

<u>= 5 M or 5 moles/liter</u>

<h3><u>Explanation;</u></h3>

At point E, 90 g of substances X are dissolved in 100 g of the solvent.

100g of the solvent is equal to 100 ml

Molarity is the number of moles of a substance in one liter of a solvent.

90 g of X are in 100 ml

But; the RFM of X = 180 g/l

Therefore; the moles of X in 90 g = 90/180

                                                        = 0.5 moles

Therefore;

0.5 moles of X are contained in 100 ml of the solvent;

Thus, molarity = 0.5 × 1000/100

                       =<u> 5 M or 5 moles/liter</u>

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Determine the rate of a reaction that follows the rate law:
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Answer:

k= 1.5

[A] = 1 M

[B] = 3 M

m = 2

n = 1

Explanation:

rate = k[A]”[B]"

6 0
2 years ago
Hno3, a strong acid, is added to shift the ag2co3 equilibrium (equation 7.6) to the right. explain why the shift occurs.
Deffense [45]
Answer 1)  When a strong acid like HNO_{3} reacts with Ag_{2} CO_{3} usually the equilibrium shifts to the right because

As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change."  So, in the given reaction when HNO_{3} reacts with Ag_{2} CO_{3}  it generates carbon dioxide and water as a by product, if we are adding HNO_{3} it will remove some of the CO_{3} molecule from the reaction mixture, which then tends to shift the equilibrium towards right.

Answer 2) The same would be observed in this case, if we replace HNO_{3} with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.

As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing Ag^{+} and Cl^{-} ions from the solution.
7 0
2 years ago
Which type of bonding is most important in ch3ch2ch2ch2ch2ch3?
saveliy_v [14]

The compound CH_3CH_2CH_2CH_2CH_2CH_3 is formed only by sharing of electrons between the atoms. The structure of the compound is shown in the image.

Each line between two atoms represents the sharing of an electron pair which results in the formation of a single bond. Since, carbon has 4 electrons in its valence shell and hydrogen has 1 electron in its valence shell so in order to complete the octet ( to have 8 electrons in their valence shell, noble gas configuration) to attain stability carbon needs 4 more electrons and hydrogen needs 1 electron. So, sharing of electron will occur as shown in the image and the formed compound is stable in nature.

Since, the bond that is formed by sharing of electrons between atoms is known as covalent bond. So, covalent bonding is most important in CH_3CH_2CH_2CH_2CH_2CH_3.

7 0
3 years ago
Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expres
Marizza181 [45]

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

Ksp=[Ag^+][Cl^-]

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

       \ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)

I          -                   0             0

C        -                   +x           +x

E        -                    x             x

Which leads to the following modified equilibrium expression:

Ksp=x^2

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!

6 0
2 years ago
Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3
maks197457 [2]
To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


4 0
3 years ago
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