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Oxana [17]
3 years ago
11

What must be true for a substance to be able to dissolve in water?

Chemistry
1 answer:
JulsSmile [24]3 years ago
6 0
B. i think not 100% sure 
You might be interested in
f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration
goldfiish [28.3K]

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

7 0
3 years ago
Why is Anton Von Leeuwenhoek important in the cell theory?​
vesna_86 [32]

Answer:

He was the first scientist to observe and describe bacteria and protozoa by looking at a drop of water from a pound under a microscope. He also was the one to build the first compound microscope.

Hope this helps :)

4 0
3 years ago
Some amount of hydrogen peroxide (H2O2) breaks down to produce 3 molecules of oxygen (O2) and 6 molecules of water (H2O). How ma
soldier1979 [14.2K]

Answer:

Atoms_H=12Atoms

Explanation:

Hello,

In this case, the only source of hydrogen is in the 6 molecules of water, therefore, the atoms of hydrogen, by applying stoichiometry with the Avogadro's number is:

Atoms_H=6moleculesH_2O*\frac{1molH_2O}{6.022x10^{23}moleculesH_2O}\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}AtomsH}{1molH} \\Atoms_H=12Atoms

Best regards.

3 0
2 years ago
Read 2 more answers
An acetic acid buffer solution is required to have a pH of 5.27. You have a solution that contains 0.01 mol of acetic acid. What
Kipish [7]

Answer:

the molarity of sodium acetate is 53

7 0
2 years ago
()
const2013 [10]

Answer:

\large \boxed{\text{-1276 kJ/mol}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                             CH₃CH₂OH        +  3O₂ ⟶ 2CO₂ + 3H₂O

Bonds:         5C-H 1C-C 1C-O 1O-H    3O=O     4C=O   6O-H

D/kJ·mol⁻¹:    413    347  358  467       495        799      467

\Delta H = \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\\sum{D_{\text{reactants}}} = 5 \times 413 + 1 \times 347 + 1 \times 358 + 1 \times 467 + 3 \times 495 = 3237 + 1485\\=\text{4722 kJ}\\\sum{D_{\text{products}}} = 4 \times 799 + 6 \times 467 =3196 + 2802 = \text{5998 kJ}\\\Delta H = 4722 - 5998= \textbf{-1276 kJ} \\ \text{The overall energy change is $\large \boxed{\textbf{-1276 kJ/mol}}$}.

6 0
3 years ago
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