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IceJOKER [234]
3 years ago
15

A 0.8115 g sample of HCl was placed into a 50 mL volumetric flask and the sample was thoroughly dissolved in water to make 50 mL

of solution. It required 22.07 mL of NaOH to reach the endpoint in the titration. What is the molarity of the NaOH solution?
Chemistry
1 answer:
inysia [295]3 years ago
5 0

Answer:

Molarity of NaOH solution is 1.009 M

Explanation:

Molar mass of HCl is 36.46 g/mol

Number moles = (mass)/(molar mass)

So, 0.8115 g of HCl = \frac{0.8115}{36.46}moles HCl = 0.02226 moles HCl

1 mol of NaOH neutralizes 1 mol of HCl.

So, if molarity of NaOH solution is S(M) then moles of NaOH required to reach endpoint is \frac{S\times 22.07}{1000}moles

So, \frac{S\times 22.07}{1000}=0.02226

or, S = 1.009

So, molarity of NaOH solution is 1.009 M

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

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Petrol is a conductor of electricity, because it conducts rather than insulates. 
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How many grams of butane can be burned by 1.42 moles of oxygen?
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Answer:

So the molar mass of C4,H10 is

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why is it harder to remove an electron from fluorine than from carbon? to put it another way, why are the outermost electrons of
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It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.

The energy required to remove an electron from an atom is called ionization energy.

The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.

Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that;  it is harder to remove an electron from fluorine than from carbon.

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3 years ago
If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of
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This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

M=\frac{n}{V}

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

2.60g*\frac{1mol}{102.89g} =0.0253mol

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

M=\frac{0.0253mol}{0.160L}=0.158M

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L

That in milliliters turns out to be:

V=0.211L*\frac{1000mL}{1L}=211mL

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2 years ago
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