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Nimfa-mama [501]
3 years ago
6

The reaction of sulfuric acid (H2SO4) with potassium hydroxide (KOH) is described by the equation below. Suppose 0.05 L of KOH w

ith unknown concentration is placed in a flask with bromthymol blue indicator. A solution of 0.10 M H2SO4 is dripped into the KOH solution. After exactly 0.033 L of H2SO4 is added, the indicator changes from blue to yellow. What is the concentration of the KOH.
H2SO4+2KOH--->K2SO4+2H2O
Chemistry
2 answers:
Dmitry_Shevchenko [17]3 years ago
7 0

chem eqn is H2SO4+2KOH--->K2SO4+2H2O

1x H2SO4 reacts with 2x KOH

0.033L of 0.10M H2SO4 has 0.033 x 0.10 = 0.0033 moles

2x 0.0033 = 0.0066 moles of KOH

conc of KOH = moles/vol = 0.0066/0.05

=0.132M

strojnjashka [21]3 years ago
3 0

From the chemical equation given:

H2SO4+2KOH--->K2SO4+2H2O

the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.

No. of moles of KOH = 2* no. of moles of H2SO4

=2*0.1*0.033

The concentration of KOH = no. of moles / volume

=2*0.1*0.033/0.05

=0.132M

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Explain why beryllium looses electrons while ionic bonds,while Sulfur gains electrons.
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Answer:

Since Beryllium has a larger atomic radius than Sulphur its electrons are not strongly attracted to the nucleus hence lost easily. But Sulphur has a small atomic radius hence electrons are more closely attracted to the nucleus.

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1 year ago
How many grarns would 3 moles of carbon dioxide (CO2) weigh? (round to the
BigorU [14]

Mass of CO₂ = 132 g

<h3>Further explanation   </h3>

A mole is a number of particles(atoms, molecules, ions)  in a substance

This refers to the atomic total of the 12 gr C-12  which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mole also can be formulated :

\tt n=\dfrac{mass}{MW}

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6 0
3 years ago
Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
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Answer:

0.269 g

Explanation:

Po_{210} ^{84}  ⟶  Pb_{206} ^{82}+  + He_{4} ^{2}

Data:

Half-life of  Po_{210} ^{84}  (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol  

Time = 338.8 days  

Isotopic masses  

Po_{210} ^{84} = 209.98 g/mol  

Pb_{206} ^{82} = 205.97 g/mol  

Concepts  

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.  

The radioactive decay law is  

N=Noe^(-λt)

Where:  

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

λ= radioactive decay constant  

The radioactive constant is related to the half-life by the next equation  

λ= \frac{ln 2}{t(1/2)}

so  

λ= \frac{ln2}{138.4 days}  =0,005008 days^(-1)

No (Atoms of  Po_{210} ^{84}  in t=0)

To get No we need to calculate the number of atoms of  Po_{210} ^{84}   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  Po_{210} ^{84}  in the initial sample.  

This is:

\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol} = 0,001599 mol of  PoCl4

This is the number of mol of  Po_{210} ^{84} in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  Po_{210} ^{84} * 6.023*10^23 atoms/ mol of  Po_{210} ^{84} = 9.632 *10^20 atoms of  Po_{210} ^{84}

Atoms after 338.8 days

We use the radioactive decay law to get this value  

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of  Po_{210} ^{84} in the sample after 338.8 days has passed  

The number of atoms  Po_{210} ^{84} transformed is equal to the number of atoms of Pb_{206} ^{82}  produced.  

The number of atoms of Po_{210} ^{84} transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of Pb_{206} ^{82} produced  

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of Pb_{206} ^{82} ) / ( 6.023*10^23 atoms of Pb_{206} ^{82} / mol of Pb_{206} ^{82} =  0.001306 moles of Pb_{206} ^{82}

This number of moles have a mass of:

(0,001306 moles of Pb_{206} ^{82} )* (205.97 g of Pb_{206} ^{82} /mol of Pb_{206} ^{82} ) = 0.269 g  

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3 years ago
Question 2 (1 point)<br>At the center of the atom is the electron.<br>True<br>False​
Stells [14]

Answer:

False the electrons are on the outside of the atoms

Explanation:

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3 years ago
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