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Arturiano [62]
3 years ago
14

If you have 40.0 g HCl and 40.0 g NaHCO3, do you have the same number of moles of each?

Chemistry
1 answer:
umka2103 [35]3 years ago
3 0

They both have a different number of moles

<h3>Further explanation</h3>

Given

40 g HCl

40 g NaHCO3

Required

moles

Solution

The mole is the number of particles contained in a substance

1 mol = 6.02.10²³

Moles can also be determined from the amount of substance mass and its molar mass

n = mass : molar mass

So the moles of a compound are affected by its molar mass (inversely proportional)

The larger the molar mass (of the same mass), the smaller the moles

molar mass HCl = 58.5 g/mol

molar mass NaHCO3 = 84 g/mol

mol HCl :

= 40 : 58.5

= 0.684

mol NaHCO# :

= 40 : 84

= 0.476

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Consider NH3 and PH3. Electronegativities: P = 2.1, H = 2.1, N=3.0. Which statement is false?
NikAS [45]

Answer:

The P-H bonds are more polar than the N-H bonds.

Explanation:

Phosphine is a polar molecule with non-polar bonds. The phosphorus atom is bonded to three hydrogen atoms and the phosphorus atom has a lone pair of electrons. Since hydrogen and phosphorus are equal in electronegativity, it implies that they attract the shared pairs of electrons the same amount,hence bonding electrons are shared equally making the covalent bonds non-polar.

The lone pair of electrons on phosphorus causes the molecule to be asymmetrical with respect to charge distribution this is why the molecule is polar even though the are non-polar bonds in the molecule.

Looking at the values of electro negativity stated in the question, one can easily see that the difference in electro negativity between nitrogen and hydrogen is 0.9 while the difference in electro negativity between phosphorus and hydrogen is zero. It is clear that NH3 is naturally more polar than PH3 since each individual N-H bond in NH3 is a polar bond while the individual P-H bonds in PH3 are nonpolar.

4 0
3 years ago
Now, Johnny the pool guy must add enough Na2CO3 to get the pH back up to 7.60. How many grams of Na2CO3 does Johnny need to add
vaieri [72.5K]

Answer:

m = 3.4126 g

Explanation:

First, the question is incomplete but I already put in the comments the rest of the question.

Let's solve the first two questions, and then the actual question you are asking here to give you a better explanation of how to do it.

1) We need the volume of the pool, in this case is easy. Assuming the pool is rectangular, we use the volume of a parallelepiped which is the following:

V = h * d * w

We have the data, but first we will convert the feet to centimeter. This is because is easier to work the volume in cm³ than in feet.

So the height, width and depth of the pool in centimeter are:

h = 32 * 30.48 = 975.36 cm

w = 18 * 30.48 = 548.64 cm

d = 5.3 * 30.48 = 161.54 cm

Now the volume:

V = 975.36 * 548.64 * 161.54

V = 86,443,528.79 cm³ or 86,443,528.79 mL or 86,443.5 L

2) If the pool has a pH of 6.4, the concentration of H+ can be calculated with the following expression:

[H+] = antlog(-pH) or 10^(-pH)

Replacing we have:

[H+] = 10^(-6.4)

[H+] = 3.98x10^-7 M

3) Finally the question you are asking for.

According to the reaction:

Na2CO3 + H+ → 2Na+ + HCO3−

We can see that there is ratio of 1:1 between the H+ and the Na2CO3, so, if we have initially a concentration of 3.98x10^-7 M, the difference between the new concentration of H+ and the innitial, will give the concentration to be added to the pool to raise the pH. Then, with the molecular weight of Na2CO3 (105.98 g/mol) we can know the mass needed.

The new concentration of [H+] is:

[H+] = 10^(-7.6) = 2.58x10^-8 M

The difference of both [H+] will give concentration of Na2CO3 used:

3.98x10^-7 - 2.58x10^-8 = 3.73x10^-7 M

The moles:

moles = 3.73x10^-7 * 86,443.5 = 0.0322 moles

Finally the mass:

m = 0.0322 * 105.98

m = 3.4126 g    

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