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Ber [7]
4 years ago
6

Pure potassium hydrogen phthalate is used for the standardization of the sodium hydroxide solution. Suppose that the potassium h

ydrogen phthalate is not completely dry. Will the reported molar concentration of the sodium hydroxide solution be too high, too low, or unaffected because of the moistness of the potassium hydrogen phthalate?
Chemistry
1 answer:
VARVARA [1.3K]4 years ago
6 0

Answer:

Too high

Explanation:

The standarization of sodium hydroxide solution with potassium hydrogen phthalate consist in the addition of known quantity of moles of acid by weighing this solid.

If the solid is moist, the real mass of the acid will be less than weighed doing less the real moles of potassium hydrogen phthalate. Thus, in titration, volume of NaOH will be less than real volume and reported molar concentration will <em>be too high.</em>

<em />

I hope it helps!

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Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
elixir [45]

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  

3 0
3 years ago
Characteristics of a medium wave
ehidna [41]
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4 0
3 years ago
Read 2 more answers
Is KNO a strong acid?
stira [4]

Answer:

First of all, it's KNO₃ not KNO.

Second, KNO₃ is neither an acid nor it is a base, infact, it is a salt and therefore it's neutral.

hope that helps...

3 0
3 years ago
Please need this ASAP. Calculate the mass of lime, CaO, that would be produced from 250 tonnes of limestone,
stiv31 [10]

Answer:

1.4×10⁸ g of CaO

Explanation:

We'll begin by converting 250 tonnes to grams (g). This can be obtained as follow:

1 tonne = 1×10⁶ g

Therefore,

250 tonne = 250 × 1×10⁶

250 tonne = 2.5×10⁸ g

Next, the balanced equation for the reaction.

CaCO₃ —> CaO + CO₂

Next, we shall determine the mass of CaCO₃ that decomposed and the mass CaO produced from the balanced equation. This can be obtained as follow:

Molar mass of CaCO₃ = 40 + 12 + (16×3)

= 40 + 12 + 48

= 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

Molar mass of CaO = 40 + 16

= 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

SUMMARY:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Finally, we shall determine the mass of CaO produced by the decomposition of 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃. This can be obtained as follow:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Therefore, 2.5×10⁸ g of CaCO₃ will decompose to produce =

(2.5×10⁸ × 56)/100 = 1.4×10⁸ g of CaO.

Thus, 1.4×10⁸ g of CaO will be obtained from 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃.

6 0
3 years ago
When water freezes into ice, sme of the properties have changed <br> What stays the same
larisa86 [58]
When water freezes into ice, some of the properties have change. What stays the same.




Answer: 1: Ice Melting is a physical change. When liquid water (H2O) freezes into a solid state (ice), it appears changed; However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen and 88.81% oxygen by mass.


Have a goood day!
5 0
3 years ago
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