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4 years ago

When a weak acid and a strong base are mixed, the resulting pH will be 7 (circle the correct answer).

Chemistry

1 answer:

timurjin [86]4 years ago

4 0

<u>Answer:</u> The correct answer is False.

<u>Explanation:</u>

There are three types of solution: acidic, basic and neutral

To determine the type of solution, we look at the pH values.

The pH range of acidic solution is 0 to 6.9
The pH range of basic solution is 7.1 to 14
The pH of neutral solution is 7.

When a strong acid and a weak base reacts, it leads to the formation of acidic solution. The pH of these solutions are found to be below 7.

When a strong base and weak acid reacts, it leads to the formation of basic solution. The pH of these solutions are found to be above 7.

When a strong acid and strong base or weak acid and weak base reacts, it leads to the formation of neutral solutions. The pH of these solutions are found to be equal to 7.

Hence, the correct answer is False.

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What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

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Answer:

$V=27992L=28.00m^3$

Explanation:

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In this case, the combustion of methane is shown below:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

$n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4$

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3$

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(a) Draw the best Lewis structure(s) for the nitrite ion, NO2-. (b) With what allotrope of oxygen is it isoelectronic? (c) What

maria [59]

Answer:

(a) $NO_{2}^{-}$ ion has two resonance structures.

(b) $O_{3}$ molecule is isoelectronic.

Explanation:

(a)

The Lewis structure of $NO_{2}^{-}$ is as follows.

(In attachment)

(b)

Isoelectronic means both species having same number of electrons .

$NO_{2}^{-}$ is isoelectronic with $O_{3}$ .

Both these molecules having 24 electrons and 18 valence electrons.

Hence, $O_{3}$ molecule isoelectronic with $NO_{2}^{-}$ .

(c)

The bond length of N-O single bond is more than the N-O double bond.

Due to the resonance structure of $NO_{2}^{-}$ N-O bond has double bond character.

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3 years ago

Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun

Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂= $\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}$ =0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

$\frac{0.92678}{0.0309}$ =Si³⁰

Si³⁰=29.99288

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4 years ago