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klemol [59]
3 years ago
15

If sodium sulfate was added to the mixture containing silver ions and the yellow precipitate, what might be observed after a sig

nificant amount of time elapsed?
Chemistry
1 answer:
anastassius [24]3 years ago
8 0

Explanation:

Very little to no formation of silver sulfate, as the silver sulfate Ksp( solubility product) is very large compared to the yellow precipitate Ksp.

By adding sodium sulfate to the silver / yellow precipitate mix, this question asks us to find a potential result and justification for the outcome. This implies that we will have to select the right answer that has a valid result and an adequate explanation for it. A substance's Ksp, or solubility product, is defined as the product of the dissolved ion concentration of each substance elevated to the power of its stoichiometric coefficient.

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How many moles of zinc, Zn, are in 1.31×1024 Zn atoms?
liubo4ka [24]

Answer:

=7.89013× 10^47 moles of zinc

Explanation:

1 atom contains 6.023×10^23 moles.

1.31×10^24 atoms of zinc contain6.023×10^24×1.31×10^.24

6 0
2 years ago
The actet rule indicates that
Nimfa-mama [501]

an atom will gain or lose electrons to obtain 8 electrons in it's outer shell. In other words, elements want to reach an electron configuration of the nearest noble gas.

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3 years ago
β‑Galactosidase (β‑gal) is a hydrolase enzyme that catalyzes the hydrolysis of β‑galactosides into monosaccharides. A 0.387 g sa
gtnhenbr [62]

Answer:

The molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=icRT

where,

\pi = osmotic pressure of the solution = 0.602 mbar = 0.000602 bar

0.000602 bar = 0.000594 atm

(1 atm = 1.01325 bar)

i = Van't hoff factor = 1 (for non-electrolytes)

c = concentration of solute = ?

R = Gas constant = 0.0820\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273.15 +25]=298.15 K

Putting values in above equation, we get:

0.000594 atm=1\times c\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298.15 K\\\\c=2.4278\times 10^{-5} mol/L

The concentration of solute is 2.4278\times 10^{-5} mol/L

Volume of the solution = V =0.137 L

Moles of β‑Galactosidase = n

C=\frac{n}{V(L)}

n=2.4278\times 10^{-5} mol/L\times 0.137 L

n=3.3261\times 10^{-6} mol

To calculate the molecular mass of solute, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of β‑Galactosidase = 3.3261\times 10^{-6} mol

Given mass of β‑Galactosidase= 0.387 g

Putting values in above equation, we get:

3.3261\times 10^{-6} mol =\frac{0.387 g}{\text{Molar mass of solute}}\\\\\text{Molar mass of solute}=116,352.97 g/mol

Hence, the molar mass of unknown β‑Galactosidaseis 116,352.97 g/mol.

3 0
2 years ago
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Yanka [14]

Answer:

Ag+ = 47 electrons - 1 electron = 46 electrons. Finally: 47 protons , 61 neutrons and 46 electrons.

Explanation:

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