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IrinaVladis [17]
3 years ago
8

The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions

Chemistry
1 answer:
musickatia [10]3 years ago
3 0

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,  

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,  

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.  

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,  

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,  

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,  

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,  

x as 0.115 and y as 0.165

Now the mass of AgCl will be,  

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,  

278.1 × 0.165 = 45.88 grams.  

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liubo4ka [24]

The given alkyne is Option A 3-heptyne

<h3>What is an Alkyne ?</h3>

The hydrocarbon having at least one C-C triple bond is called an Alkyne.

It has the general formula of   \rm C_n H_{2n+2} .

In the question it is being mentioned that it is an alkyne so there will be a triple bond and not a double bond.

It has been asked in the question that

CH3CH₂C ≡  CCH₂CH₂CH3 is which alkyne from the given option.

The counting of the Carbon chain is done from the left side and the Triple bond is at the 3rd Carbon , so 3-heptyne .

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2 years ago
2. Calculate the atomic mass of an element that has two isotopes, each with 50.00% abundance. One isotope has a mass of 63.00 am
melamori03 [73]

Answer:

The atomic mass of element is 65.5 amu.

Explanation:

Given data:

Abundance of X-63 = 50.000%

Atomic mass of  X-63 = 63.00 amu

Atomic mass of X-68 = 68.00 amu

Atomic mass of element = ?

Solution:

Abundance of X-68 = 100-50 = 50%

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (50×63)+(50×68) /100

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A weather balloon contains 1.10X10 to the power of 5 mol of helium and has a volume of 2.70x10 to the power of 6 L at 1.00 atm p
olchik [2.2K]

Answer:

299.14 K or 26°C

Explanation:

The ideal gas law, also called the general gas equation, is the equation of state of a hypothetical ideal gas.

The ideal gas law is often written as

PV = nRT

where P ,V and T are the pressure, volume and absolute temperature;

n is the number of moles of gas and  R is the ideal gas constant.

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when the formula is  rearranged, T=PV/ nR  

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T = 299.1421917 K

or

T = 299.14 - 273.15 = 25.99 =  26°C

4 0
3 years ago
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