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30.03.2023

How i can find the valence factor of C7H5O2Na

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29.10.2022, solved by verified expert

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Valence factor of acids , bases and salts is different.

For an acid it is no. Of replaceable H+ ions.

For base it is the No. If OH- ions.

For salts it is the valency of any cation or anion present In it.

In the question the compound given s sodium benzoate which is a salt.

C6H5COONa

Which on dissociation gives :-

C6H5COONa -------> C6H5COO- + Na+

Hence here Valence factor =1

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Chemistry

Which equivalence factor should you use to convert from 106 grams of diatomic oxygen (o2) to moles of o2? a (1 mol o2/16.00 grams o2) b (16.00 grams o2/1 mol o2) c (1 mol o2/32.00 grams o2) d (32.00 grams o2/1 mol o2)

Step-by-step answer

P Answered by Master

Answer : The correct option is, (C) $\frac{1\text{ mol }O_2}{32.00\text{ grams }O_2}$

Explanation :

As we know that, the molar mass of O_2 is 32.00 gram per mole (32 g/mole).

To convert gram into mole, we use the conversion:

We are given the mass of O_2 is 106 grams.

Converting this into mole, we get:

As, 32 gram of O_2 present in 1 mole of O_2

And, 1 gram of O_2 present in $\frac{1\text{ mol}O_2}{32.00\text{ grams}O_2}$

So, 106 gram of O_2 present in $\frac{1\text{ mol}O_2}{32.00\text{ grams}O_2}\times 106\text{ gram }O_2=3.3125\text{ mole}$

The equivalence factor use to convert gram to mole is, $\frac{1\text{ mol}O_2}{32.00\text{ grams}O_2}$

Hence, correct option is, (C) $\frac{1\text{ mol }O_2}{32.00\text{ grams }O_2}$

Chemistry

Which equivalence factor should you use to convert from 4.25 moles of al to atoms of al? (1 atom al/4.25 mol al) (6.02 x 1023 atoms al/1 mol al) (6.02 x 1023 atoms al/4.25 mol) (4.25 mol al/6.02 x 1023 atoms al)

Step-by-step answer

P Answered by PhD

3

Avogadro's number is 1 mol = 6.02 * 10^23 elements

It means that 1 mol of atoms is 6.02 * 10^23 atoms

1 mol of atoms = 6.02 * 10^23 atoms

From there, if you divide both sides by 1 mol of atoms, you get

1 = 6.02 * 10^23 atoms / 1 mol of atoms.

That means, that to pass from a number of moles of atoms to number of atoms you have to multipby by the conversion factor

6.02*10^23 atoms Al/ 1 mol Al

That is the second option of the list.

Chemistry

Which equivalence factor should you use to convert from 4.28 x 1019 molecules of oxygen (o2) to moles of o2? (1 mol o2/4.28 x 1019 molecules o2) (4.28 x 1019 molecules o2/1 mol o2) (1 mol o2/6.02 x 1023 molecules o2) (6.02 x 1023 molecules o2/1 mol o2)

Step-by-step answer

P Answered by Specialist

1

The answer is: [C]: " (1 mol O₂/6.02 × 10²³ molecules O₂) " .

Chemistry

Which equivalence factor should you use to convert from 4.25 moles of al to atoms of al? (1 atom al/4.25 mol al) (6.02 x 1023 atoms al/1 mol al) (6.02 x 1023 atoms al/4.25 mol) (4.25 mol al/6.02 x 1023 atoms al)

Step-by-step answer

P Answered by PhD

3

Avogadro's number is 1 mol = 6.02 * 10^23 elements

It means that 1 mol of atoms is 6.02 * 10^23 atoms

1 mol of atoms = 6.02 * 10^23 atoms

From there, if you divide both sides by 1 mol of atoms, you get

1 = 6.02 * 10^23 atoms / 1 mol of atoms.

That means, that to pass from a number of moles of atoms to number of atoms you have to multipby by the conversion factor

6.02*10^23 atoms Al/ 1 mol Al

That is the second option of the list.

Chemistry

Which equivalence factor set should you use to convert 2.68 x 1011 atoms of ag to grams of ag? (2.68 x 1011 atoms ag/1 mol ag)(1 mol ag/107.88 g ag) (1 mol ag/2.68 x 1011 atoms ag)(107.88 g ag/1 mol ag) (1 mol ag/6.02 x 1023 atoms ag)(107.88 g ag/1 mol ag) (2.68 x 1011 atoms ag/6.02 x 1023 atoms ag)(107.88 g ag)

Step-by-step answer

P Answered by PhD

$(2.68\times 10^{11}atoms Ag/6.02\times 10^{23}atoms Ag)(107.88 g Ag)$

Explanation:

According to Avogadro's law, 1 mole of every substance contains avogadro's number $(6.023\times 10^{23})$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given atoms}}{\text {Avogadro's number}}$

For Ag

given atoms = $2.68\times 10^{11}$

avogadro's number = $6.023\times 10^{23}$

Putting values in above equation, we get:

$\text{Moles of}Ag=\frac{2.68\times 10^{11}}{6.023\times 10^{23}}$

1 mole of Ag weighs = 107.88 grams

Thus $\frac{2.68\times 10^{11}}{6.023\times 10^{23}}$ moles of Ag will weigh= $\frac{107.88}{1}\times \frac{2.68\times 10^{11}}{6.023\times 10^{23}}=43.2\times 10^{-12}$ grams.

Chemistry

Which equivalence factor set should you use to convert from 4.23 x 1012 atoms of ca to grams of ca? a) (1 mol ca/6.02 x 1023 atoms ca)(40.08 g ca/1 mol ca) b) (1 mol ca/4.23 x 1012 atoms ca)(40.08 g ca/1 mol ca) c) (4.23 x 1012 atoms ca/1 mol ca)(1 mol ca/40.08 g ca) d) 4.23 x 1012 atoms ca/6.02 x 1023 atoms ca)(40.08 g ca)

Step-by-step answer

P Answered by Specialist

3

$\frac{4.23 * 10^{12} atoms Ca}{6.02 * 10^{23} atoms Ca}$ * 40.08 g Ca

Explanation:
1- One mole of any substance contains Avogadro's number of atoms. Avogadro's number = 6.02 * 10²³
To get the number of moles is 4.23 * 10¹² atoms, we will use cross multiplication as follows:
1 mole .................> 6.02 * 10²³ atoms
x moles ................> 4.23 * 10¹²
x = (4.23 * 10¹² atoms Ca) / (6.02 * 10²³ atoms Ca) moles

2- Number of moles can be calculated as follows:
number of moles = mass / molar mass
We know that the molar mass of Ca = 40.08 g/mol, therefore:
number of moles of Ca = mass / 40.08
This means that:
mass of Ca = number of moles of Ca * 40.08

Substitute with the number of moles (x) calculated in part 1 in the above equation to get the mass as follows:
mass of Ca = $\frac{4.23 * 10^{12} atoms Ca}{6.02 * 10^{23} atoms Ca}$ * 40.08 g Ca

Hope this helps :)

Chemistry

Which equivalence factor set should you use to convert 126 g c to atoms c? (1 mol c/126 g c)(6.02 x 1023 atoms c/1 mol c.(1 mol c/12.01 g c)(6.02 x 1023 atoms c/1 mol c.(12.01 g c/1 mol c)(1 mol c/6.02 x 1023 atoms c.(12.01 g c/1 mol c)(6.02 x 1023 atoms c/1 mol c)?

Step-by-step answer

P Answered by Master

3

For this you must Cross multiply and cancel any two units that are the same and diagonal from each other.

I attached a screenshot of my work. Hopefully it all makes sense.
Which equivalence factor set should you use to convert 126 g c to atoms c? (1 mol c/126 g c)(6.02 x

Which equivalence factor set should you use to convert 126 g c to atoms c? (1 mol c/126 g c)(6.02 x

Chemistry

Which equivalence factor should you use to convert from 106 grams of o2 to moles of o2? a.(1 mol o2/16.00 grams o2) b.(16.00 grams o2/1 mol o2) c.(1 mol o2/32.00 grams o2) d.(32.00 grams o2/1 mol o2) pls for exam

Step-by-step answer

P Answered by PhD

2

C. (1 mol O2/32.00 g O2)

To convert 106 g O2 to moles of O2, the equivalence factor is either

(1 mol O2/32.00 g O2) or (32.00 g O2/1 mol O2). We choose the one that has the desired units (mol) on top: (1 mol O2/32.00 g O2). Then,

Mass of O2= 106 g O2× (1 mol O2/32.00 g O2) = 3.31 mol O2

Notice that the unwanted units (g O2) cancel and do not appear in the answer.

“A. (1 mol O2/16.00 g O2)” is incorrect because it uses the molar mass of O (not O2).

”B. (16.00 grams O2/1 mol O2) is incorrect because it uses the wrong molar mass for O2.

“D. (32.00 g O2/1 mol O2)” is incorrect because it has the equivalence factor upside down.

Chemistry

Which equivalence factor should you use to convert from 38.2 grams of k to moles of k? a (38.2 g k/39.10 g k) b (38.2 g k/1 mol k) c (1 mol k/38.2 g k) d (1 mol k/39.10 g k)

Step-by-step answer

P Answered by PhD

(38.2 g K/39.10 g K)

Explanation:

To convert given masses into moles we use the formula :

Moles of K= $\frac{\text{ given mass of K}}{\text{ molar mass of K}}$

Given:

Mass of K= 38.2 grams

Molar mass of K= 39.10 g/mol

Putting in the values in the given formula, we get:

Moles of K= $\frac{38.2g}{39.10g/mole}=0.98moles$

Thus the moles are 0.98 moles.

Chemistry

Which equivalence factor should you use to convert from 4.28 x 1019 molecules of oxygen (o2) to moles of o2? (1 mol o2/4.28 x 1019 molecules o2) (4.28 x 1019 molecules o2/1 mol o2) (1 mol o2/6.02 x 1023 molecules o2) (6.02 x 1023 molecules o2/1 mol o2)

Step-by-step answer

P Answered by Specialist

1

The answer is: [C]: " (1 mol O₂/6.02 × 10²³ molecules O₂) " .

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