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anyanavicka [17]
3 years ago
11

Solute: Will give brainliest

Chemistry
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:

58.45 g of NaCI

Explanation:

hope this helps ya

have a great day

brainliest if its right?

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What is the difference between glucose and fructose.
CaHeK987 [17]

Answer:

glucose is in large sugar molecules, while fructose is in simple fruits and vegetables

4 0
2 years ago
Read 2 more answers
How are acids and ionic compounds similar?
11Alexandr11 [23.1K]

Answer: Ionic compounds are held together by the virtue of their opposing charges. Na+Cl- for example. If we consider Hg+(2Cl-)2, a mercuric chloride, the solubility is much less. Ba++(SO)4 Barium Sulphate, is highly insoluble; all differ by the relative attractiveness by Differing opposing charge(s).

Acids are very similar, consider Formic Acid, HCOOH, the simplest of the Carboxylic Acids. It dissociates more than say Benzoic Acid, C6H5-COOH. But neither disassociate as fully as Nitric Acid HNO3.

So the relative disassociation of the H+ (proton), or H3O+, (Hydronium ion), from any of these in water vary for a number of reasons we need not consider now.

Here is a “Tricky One!” (And very nasty). Take HF liquid or gas. This is one of the strongest acids on Earth - AS A LIQUID compound OR GAS. It will dissociate essentially near completion! Eat the floor, and is very dangerous.

NOW - HF (aqueous). The HF is in water. Very like HCl? NO! Why you may ask...The Electrophilic nature of Fluorine, “bathed in water, with an H+ all its own”, doesn’t let it go as easily!

HF is HIGHLY ordered in water, you can almost imagine a sort of “Hydrated matrix”, little HFs in endless rows...

BUT BE WARNED - even the aqueous HF is so reactive it will dissolve bone!

(I was told it was extremely painful; and did not appear to heal for weeks!)

Explanation: so, both types of compounds have a similarity, held together by the strength of their opposing charges or the degree of dissociation, (using water for simplicity).

That should do it.

8 0
3 years ago
consider this reaction at equilibrium at a total pressure: 2h2o(g) o2(g) 2h2o2(g) suppose the volume of this system is twice its
Daniel [21]

The total pressure when the new equilibrium is stabilized is half of the initial pressure of the system.

The given chemical reaction at a stable equilibrium is,

2H₂O(g)+O₂(g) = 2H₂O₂(g)

According to the ideal gas equation,

PV = nRT

P is pressure,

V is volume,

n is moles

R is gas constant,

T is temperature.

Assuming the temperature is constant.

If the volume of the system is twice the initial volume then the total pressure at the new equilibrium can be found out as,

P₁V₁ = P₂V₂

Where, P₁ and V₁ are initial volume and pressure while P₂ and V₂ are final pressure and volume.

If V₂ = 2V₁,

P₂ = P₁/2

So, the final total pressure will be half of the initial pressure.

To know more about equilibrium, visit,

brainly.com/question/517289

#SPJ4

5 0
1 year ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

7 0
3 years ago
Which term describes rocks that absorb water?
KIM [24]

Answer:

C. porous

hope it helps!!!

please mark as the brainliest if it is correct!

8 0
3 years ago
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