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r-ruslan [8.4K]
3 years ago
11

Short, hairlike appendage

Chemistry
1 answer:
LekaFEV [45]3 years ago
6 0
Cilium part of a cell
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The particles remain seperate and don’t combine? What is this a mixture or a solution?
djyliett [7]
It is actually a Colloid, which means it will never mix and settle.<span />
3 0
3 years ago
The density of ethanol is 0.788 g/mL. What is the mass of 157 mL of alcohol?
Lina20 [59]

Answer:

<h3>The answer is 123.72 g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 157 mL

density = 0.788 g/mL

We have

mass = 0.788 × 157 = 123.716

We have the final answer as

<h3>123.72 g</h3>

Hope this helps you

6 0
2 years ago
Which of these best describes a scientific model?
goblinko [34]
The answer a way of explaining a complex concept.
3 0
3 years ago
Calculate molality,molarity and mole fraction of KI if the density of 20%(mass) aqueous KI is 1.202 g/mL
Scorpion4ik [409]
Basis: 1 L of the substance.
               (1.202 g/mL) x (1000 mL) = 1202 g 
                   mass solute = (1202 g) x 0.2 = 240.2 g
                   mass solvent = 1202 g x 0.8 =  961.6 g
                   moles KI = (240.2 g) x (1 mole / 166 g)  = 1.45 moles
                   moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent 
                  = 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
                  = 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
                        =  1.45 moles / (1.45 moles + 53.42 moles) = 0.0264 

5 0
3 years ago
50 ml of 0.2 M acetic acid is shaken with 10 g activated charcoal. The concentration of acetic acid is reduced to 0.5 times the
Olegator [25]

Answer:

0.03 g_{acid}/g_{charcoal}

Explanation:

The amount adsorbed (solute) is the acetic acid, and the adsorbent is the activated charcoal. The mass of the adsorbent is 10 g.

So, we need to calculate the mass of the acetic acid as follows:

m = n*M = C*V*M

Where:

n: is the number of moles = C*V

M: is the molecular mass =  60.052 g/mol

C: is the final concentration of the acid = 0.5*0.2 mol/L = 0.10 mol/L

V: is the volume = 50 ml = 0.050 L

m_{acid} = C*V*M = 0.10 mol/L*0.050 L*60.052 g/mol = 0.30 g

Now, the amount of solute adsorbed per gram of the adsorbent is:

\frac{m_{acid}}{m_{charcoal}} = \frac{0.30 g}{10 g} = 0.03 g_{acid}/g_{charcoal}

Therefore, the amount of solute adsorbed per gram of the adsorbent is 0.03 g/g.

I hope it helps you!

3 0
2 years ago
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