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3 years ago

What is the most effective way to differentiate between potassium feldspar and plagioclase feldspar?

Chemistry

1 answer:

Tresset [83]3 years ago

3 0

<span>plagioclase feldspars have striations and potassium feldspars don't have striations</span>

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You have 500 mL of 5 M HCl already made. You need to dilute the solution to 1 M HCl. How much water will you need to add?

hram777 [196]

Answer:

You need to add 400mL of water

Explanation:

500mL = 5 M HCI That means that if you divide both sides by 5

100mL = 1 M HCI If you need ot get rid of 4 M HCI then you add 400 mL of water because that is what it is equal to

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3 years ago

Which of the following are physical properties of nonmentals

coldgirl [10]

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6 0

2 years ago

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7 0

2 years ago

Given that you started with 28.5 g of K3PO4, how many grams of KNO3 can be<br> produced?

Irina18 [472]

Mass of KNO₃ : = 40.643 g

<h3>Further explanation</h3>

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

= 6/2 x mol K₃PO₄

= 6/2 x 0.134

= 0.402

Mass of KNO₃ :

= mol x MW KNO₃

= 0.402 x 101,1032 g/mol

= 40.643 g

8 0

2 years ago

The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react

Kazeer [188]

Answer:

The percent yield of chloro-ethane in the reaction is 82.98%.

Explanation:

$C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl$

Moles of ethane = $\frac{300.0 g}{30 g/mol}=10 mol$

Moles of chlorine gases = $\frac{650.0 g}{71 .0 g/mol}=9.1549 mol$

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

$\frac{1}{1}\times 9.1549 mol=9.1549 mol$ of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

$\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%$

The percent yield of chloro-ethane in the reaction is 82.98%.

6 0

3 years ago