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ExtremeBDS [4]
3 years ago
5

A volume of 30.0 mL of a 0.140 M HNO3 solution is titrated with 0.570 M KOH. Calculate the volume of KOH required to reach the e

quivalence point.
Chemistry
1 answer:
igomit [66]3 years ago
5 0

Answer:

7.37 mL of KOH

Explanation:

So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,

HNO3 + KOH → KNO3 + H2O

Step 1 : The moles of HNO3 here can be calculated through the given molar mass (  0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.

Mol of NHO3 = 0.140 M * 30 / 1000 L = 0.140 M * 0.03 L = .0042 mol

Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,

0.0042 mol HNO2 * ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH

From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,

Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).

Volume of KOH = 0.0042 mol * ( 1 L / 0.570 mol ) * ( 1000 mL / 1 L ) = 7.37 mL of KOH

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Answer:

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3 years ago
How much energy do individual photons of 470 nm light have
IceJOKER [234]
<h3>Answer:</h3>

4.227 × 10^-19 Joules

<h3>Explanation:</h3>

Energy of a photon of light is calculated by the formula;

E = hf, where h is the plank's constant, 6.626 × 10^-34 J-s and f is the frequency.

But, f = c/λ

Where, c is the speed of light (2.998 × 10⁸ m/s), and λ is the wavelength.

Given the wavelength is 470 nm or 4.7 × 10^-7 m

Therefore;

E = hc/λ

  = (6.626 × 10^-34 J-s × 2.998 × 10^8 m/s) ÷ 4.7 × 10^-7 m

  = 4.227 × 10^-19 Joules

Therefore, the energy of a photon with 470 nm is 4.227 × 10^-19 Joules

3 0
3 years ago
he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp
wolverine [178]

1) Drift velocity: 3.32\cdot 10^{-4}m/s

2) 5.6\cdot 10^{13} electrons per person

Explanation:

1)

For a current flowing through a conductor, the drift velocity of the electrons is given by the equation:

v_d=\frac{I}{neA}

where

I is the current

n is the concentration of free electrons

e=1.6\cdot 10^{-19}C is the electron charge

A is the cross-sectional area of the wire

The cross-sectional area can be written as

A=\pi r^2

where r is the radius of the wire. So the equation becomes

v_d=\frac{I}{ne\pi r^2}

In this problem, we have:

I = 8.0 A is the current

8.5\cdot 10^{28} m^{-3} is the concentration of free electrons

d = 1.5 mm is the diameter, so the radius is

r = 1.5/2 = 0.75 mm = 0.75\cdot 10^{-3}m

Therefore, the drift velocity is:

v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s

2)

The total length of the cord in this problem is

L = 3.00 m

While the cross-sectional area is

A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2

Therefore, the volume of the cord is

V=AL (1)

The number of electrons per unit volume is n, so the total number of electrons in this cord is

N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}

In total, the Earth population consists of 8 billion people, which is

N'=8\cdot 10^9

Therefore, the number of electrons that each person would get is:

N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}

7 0
3 years ago
A 25.0 ml sample of 0.150 m butanoic acid is titrated with a 0.150 m naoh solution. what is the ph after 26.0 ml of base is adde
lesantik [10]
First,
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                        =  25ml/1000 * 0.15 = 0.00375 moles
moles of NaOH = volume per liter * moles of NaOH/liter
                           = 1 ml/1000 * 0.15 =  0.00015 moles
according to this equation:
HBu + NaOH → H2O + NaBu
when 1 mol of NaOH gives 1mol of HBu
So     0.00015 of NaOH will give 0.00015 mol of HBu 
∴moles of HBu which remains =   0.00375- 0.00015 = 0.0036 moles 

∴moles of Bu- produced = 0.00015 moles
when the total volume = 0.025 + 0.026 =0.051 L
[HBu] = 0.0036moles / 0.051 L = 0.071 moles
[Bu] = 0.00015 / 0.051L = 0.0029 moles
when Ka = [H+] [Bu] / [HBu]
1.5x10^-5 = [H+] (0.0029) /(0.071)
∴[H+] =1x10^-6 / 0.076 = 1.5 x 10^-5
∴PH = -㏒[H+]
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2 years ago
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D is the right answer

4 0
3 years ago
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