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SCORPION-xisa [38]
3 years ago
15

What are some HYDROPHILIC materials?

Chemistry
1 answer:
alisha [4.7K]3 years ago
8 0
Including the alkanes, oils , fats and greasy substances in general
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11. A 4.175 gram sample of a certain hydrate of copper (II) sulfate, CuSO,• xH,O, is heated until all
Wewaii [24]

The  formula of the hydrate = CuSO₄• 3H₂O

<h3>Further explanation</h3>

Given

4.175 grams sample CuSO₄• xH₂O

3.120 grams anhydrous compound CuSO₄

Required

The formula

Solution

mass of H₂O driven off :

= 4.175 - 3.12

= 1.055 g

MW CuSO₄ = 159.5 g/mol

MW H₂O = 18 g/mol

mol ratio of CuSO₄ : H₂O :

= 3.12/159.5 : 1.055/18

= 0.01956 : 0.05861

= 1 : 3

3 0
3 years ago
What is the role of critical thinking in scientific process
laiz [17]

Answer:

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6 0
3 years ago
Read 2 more answers
1 mole of no2(g) has a greater entropy than 1 mole of n2o4(g). true or false
bija089 [108]
In order to answer this, you need to find the empirical data for the standard entropies. Please refer to this link: http://www.mrbigler.com/misc/energy-of-formation.PDF

For NO₂ gas, the entropy is 240 J/mol-K. For N₂O₄ gas, the entropy is 304.2 J/mol-K. Therefore, <em>the statement is false.</em>
7 0
3 years ago
What is the difference between a cation and an anion?
zmey [24]
An anion has a negative charge, while a cation has a positive charge.
3 0
3 years ago
A chemist titrates 150.0 mL of a 0.2653 M carbonic acid (H2CO3) solution with 0.2196 M NaOH solution at 25 °C. Calculate the pH
xxTIMURxx [149]

Answer:

9.3

Explanation:

This is long and complicated so get ready

We are going to use the conjugate base of carbonic acid with water to make carbonic acid and OH- (Na is simply a spectator ion and is irrelavent here)

Let the conjugate base be A- and Carbonic acid be HA

A- + H20 ⇄ HA + OH-

To find the concentration of A- we must find the concentration of the reactants given. We know this will be equal because it is a strong base and all of it disassociates.

to get moles of acid we take the concentration and multiply by liters to cancel

.2653 x .150 = .039795 mol HA

Because it is at equivalence point we know the moles will be equal. We are given the concentration so we only have to solve for liters

We plug it into the equation and found: .181 L

Now use moles and combined volums to fins concentrarion which is .120 M

Now plug that use the Ka converted to Kb to find the cincentrations of HA and OH-

Ka is (10^-3.60) = 2.4E-4

Kb x Ka is 10^-14

Kb = 3.98E-11

Now we know Kb = [HA] [OH] / [A-]

Solve for this through algebra by using x for the values you dont know

youll find x^2 = 3.3E-10

X = 1.8 E -5

this is the OH- concentration

-log [oh] = pOH

pOH = 4.73

We know 14-pOH = ph so pH= 9.3

6 0
3 years ago
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