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STALIN [3.7K]
3 years ago
10

In the reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O, an equivalence point occurs when 29.43 mL of 0.1973 M NaOH is added to a 32.

42 mL portion of H2SO4. What is the [H2SO4]?
Chemistry
1 answer:
Tamiku [17]3 years ago
6 0
            moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
            moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
            [H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is  [H2SO4] = 0.08955 M

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How many total electrons can enter the set of 3p orbitals?<br> a. 3<br> b. 10<br> c. 2<br> d. 6
Sliva [168]

Answer:

For n=3 and l=1=p

It is 3p-orbital.

Magnetic quantum number m

l

have values from -l to +l and total of 2l+1 values.

Forl=1, m

l

values are:

m

l

=−1,0,1 for l=1; total m

l

values =3= Number of orbitals

Each orbital can occupy maximum of two electron

Number of electrons =2×3=6

Thus 6 electrons will show same quantum number values of n=3 and l=1.

Number of elements with last electron in 3p orbitals = 6

5 0
3 years ago
The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
The first word in the name of an ester is derived from the ________ used in the esterification.
oksian1 [2.3K]

The first word in the name of an ester is derived from the alcohol used in the esterification.

<h3>What is esterification?</h3>

Esterification is a chemical process where an organic acid with the formula is combined with an alcohol molecule having the chemical formula (ROH).

The process of esterification is known to produce an ester molecule and during this phenomenon is released water (H2O).

An example of an esterification reaction occurs when ethanoic acid (i.e., the active ingredient of vinegar) can react with C2H5OH (i.e., ethanol) in order to form the ethyl ethanoate molecule, which is a well-known ester molecule.

In conclusion, the first word in the name of an ester is derived from the Alcohol used in the esterification.

Learn more about esterification here:

brainly.com/question/14028062

#SPJ1

8 0
2 years ago
A sample of aluminum weighs 100 grams and has a volume of 37.03 cm. What is the density of
Alexxandr [17]

Answer:

2.7005 g/cm²

explanation: as we know density=mass/volume

6 0
3 years ago
Unless otherwise instructed, you may refer to the periodic table in the Chemistry: Problems and Solutions book for this question
Tom [10]
Answer: oxygen

There is the s,p,d and f blocks, from groups 1-2 that is the s block, 13-18 that’s the p block, 3-12 is the d block and the f would be lanthanide(#57-71) and actinide (#89-103).
8 0
3 years ago
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