All categories

3 years ago

what method can be used to seperate parts of liquid mixture wwhen the entire mixture can pass through a filter

1 answer:

6 0

Distillation is the <span>method that can be used to seperate parts of liquid mixture when the entire mixture can pass through a filter.</span>

You might be interested in

What was one idea Dalton taught about atoms

gogolik [260]

The main points of Dalton's atomic theory, as it eventually developed, are: Elements are made of extremely small particles called atoms. Atoms of a given element are identical in size, mass and other properties; atoms of different elements differ in size, mass and other properties.

4 0

3 years ago

Read 2 more answers

A fire women dropped a person onto the safety net.Right before the person hit the net he had a velocity of 11.2m/s and 1800 J of

Snezhnost [94]

Cheese isosisiiaiasiaiss s snananana

3 0

3 years ago

Read 2 more answers

Can someone explain to me step by step with the equation on how to do it thx

Tasya [4]

Answer:

h=1

c=12

there 8 h

then there are 5 c

Explanation:

6 0

3 years ago

Determine the maximum number of moles of product that can be produced from 7.0 mol al and 8.0 mol cl2 according to the equation

Vilka [71]

The balanced chemical reaction is

<span>2al + 3cl2 = 2alcl3

To determine the maximum amount of product, we need to determine which is the limiting reactant. Then, use the initial amount of that reactant to calculate the amount of the product that would be produced. We do as follows:

7 mol Al (3 mol Cl2 / 2 mol Al) = 10.5 mol Cl2
8 mol Cl2 ( 2 mol Al / 3 mol Cl2) = 5.3 mol Al
Thus, it is Cl2 that is the limiting reactant.

8 mol Cl2 ( 2 mol AlCl3 / 3 mol Cl2) = 5.3 moles of AlCl3 is produced</span>

5 0

3 years ago

What can you conclude about the mass of iron (III) sulfide produced when 3.0 g of iron filings react completely with 2.5 g of ye

igor_vitrenko [27]

Answer : The mass of iron(III)sulfide is, 5.4288 g

Solution : Given,

Mass of iron, Fe = 3 g

Mass of sulfur, = 2.5 g

Molar mass of Fe = 56 g/mole

Molar mass of = 256 g/mole

Molar mass of iron(III)sulfide, = 208 g/mole

The balanced chemical reaction is,

First we have to calculate the moles of iron and sulfur.

From the balanced reaction, we conclude that

16 moles of Fe react with 3 moles of

0.054 moles of Fe react with moles of

Therefore, the excess reagent in this reaction is, Fe and limiting reagent is,

Now we have to calculate the moles of FeS.

As, 3 moles of gives 8 moles of

So, 0.0098 moles of gives moles of

The moles of = 0.0261 moles

Now we have to calculate the mass of .

Mass of = Moles of × Molar mass of

Mass of = 0.0261 g × 208 g/mole = 5.4288 g

Therefore, the mass of iron(III)sulfide is, 5.4288 g

Read more on Brainly.com - brainly.com/question/6073920#readmore

5 0

3 years ago