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tankabanditka [31]
2 years ago
12

A student adds 0. 0030 mol of hcl to 100 ml of a 0. 10 m solution of a r2nh, a weak base. The ph of the solution is found to be

11. 10
Chemistry
1 answer:
egoroff_w [7]2 years ago
5 0

Since an acidic salt solution is produced when a strong acid neutralizes a weak base, the pH of the salt solution formed when HCl is added to R2NH will be less than 7.

<h3>What is a neutralization reaction?</h3>

A neutralization reaction is the react ion between an acid and a base to form salt and water only.

Neutralization reactions can either produce a neutral solution, an acidic solution or an alkaline solution at equivalence point.

When a strong acid is added to a weak base, the pH of the salt solution formed will be less than 7.

Therefore, the pH of the salt solution formed when HCl is added to R2NH will be less than 7.

Learn more about pH at: brainly.com/question/940314

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For the following, identify the most likely value for x.a. BHx b. CHx c. NHx d. CH2Clx
vekshin1

Given :

a. BH_x b. CH_x c. NH_x d. CH_2Cl_x .

To Find :

Find the most likely vale of x for each one .

Solution :

a . BH_x

Because boron have valency of 3 .

So , x = 3 .

b . CH_x

Valency of carbon is 4 .

x = 4 .

c . NH_x

Valency of nitrogen is 3 .

Therefore , x = 3 .

d . CH_2Cl_x

Now ,we know valency of carbon is 4 and hydrogen is 1 .

Also , two hydrogen are already there .

So , only 2 electrons left to share .

Since , chlorine have valency of 1 .

Therefore , only 2 electrons of chlorine can connect .

x = 2 .

Hence , this is the required solution .

5 0
3 years ago
Combustion of hydrocarbons such as butane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosp
Dahasolnce [82]

Answer:

1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. V = 596L

Explanation:

Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:

C₄H₁₀ + O₂ → CO₂ + H₂O

1. The balanced chemical equation is:

C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. 0,360kg of butane are:

360g×\frac{1mol}{58,12g}=<em>6,19moles of butane</em>

These moles of butane are:

6,19moles of butane×\frac{4CO_2}{1molButane}= <em>24,8 moles CO₂</em>

Using V=nRT/P

Where:

n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).

Volume (V) is:

<em>V = 596L</em>

I hope it helps!

8 0
3 years ago
Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine
yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

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3 years ago
How is ti possible that, when adding protons while moving left to right across a period, the size of an atom shrinks? Why is the
Kaylis [27]

As one moves across a period, from left to right, both the number of protons and electrons of a neutral atom increase. The enhancing number of electrons and protons results in a greater attraction between the electrons and the nucleus. This uplifted attraction pulls the electrons nearer to the nucleus, therefore, reducing the size of the atom.  

On the other hand, while moving down a group, there is an increase in the number of energy levels. The enhanced number of energy levels increases the size of the atom in spite of the elevation in the number of protons. In the outermost energy levels, the protons get attracted towards the nucleus, however, the attraction is less due to an increase in the distance from the nucleus.  

4 0
3 years ago
HELP WITH A THESIS STATEMENT
lina2011 [118]

The second thesis statement is perfect. It supports the claim and presents main idea.

4 0
3 years ago
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