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DIA [1.3K]
3 years ago
5

The strength of an acid is affected by the polarity of the bond connected to the acidic hydrogen. The more highly polarized this

bond, the more easily the hydrogen is ionized. Electronegative atoms or groups of atoms present in the structure of an acid can act to withdraw electrons and produce additional polarization. Two common groups of acids to which this principle can be applied are oxoacids and carboxylic acids. In the latter group, the length of the hydrocarbon chain in a carboxylic acid has very little effect on acid strength Longer chains may slightly diminish acidity. Bases act as hydrogen ion acceptors because of the unshared electron pass in their structure. Any group present in a base that withdraws electrons makes these electron pairs less available to accept a hydrogen ion. In contrast, any group that can act as an electron donating group such as hydrocarbon groups (usually represented as II) can increase the base strength. Thus, the addition of electronegative atoms or groups of atoms to the structure of a base decreases the base strength and electron donating groups increase base strength. Many common weak bases are derivatives of ammonia, in which H atom(s) of NH_2 are replaced with other groups.
Arrange the following oxoacids in order of decreasing acid strength. Rank from strongest to weakest acid.
1. HBrO
2. HClO
3. HClO2
4. HClO3
Chemistry
1 answer:
Shtirlitz [24]3 years ago
5 0

Answer:

The answer is "\bold{HClO_3 > HClO_2 >HClO > HBrO}"

Explanation:

We arrange oxoacids to decrease the intensity of acids in this question. Or we may conclude all this from strongest to weakest acids they order oxoacids, that's why above given order is correct.

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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 years ago
What is the formula for percent yield
Sindrei [870]
I think it is 89.3%?
3 0
3 years ago
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0
3 years ago
Read 2 more answers
What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i
UNO [17]

Answer:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)

The net equation is then:

Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-

Actual cathode half-equation:

Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)

Actual net reaction:

Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)

6 0
3 years ago
Which function does photosynthesis replace for plants?
murzikaleks [220]

Answer:

Breathing.

Explanation:

7 0
3 years ago
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