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yKpoI14uk [10]
3 years ago
14

Describe the two eukaryotic animal cell organelles that have roles similar to the organs of the human digestive system.

Chemistry
1 answer:
Mariulka [41]3 years ago
6 0

Two eukaryotic animal cell organelles that have roles similar to the organs of the human digestive system are lysosome and mitochondria. The lysosome is like our stomach, it is responsible for digesting or breaking down the food consumed by the cell. Mitochondria is like the small intestine, it is responsible for converting the food we eat into nutrients and minerals needed for the cell.
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What is the molar solubility in water of ag2cro4 ? (the ksp for ag2cro4 is 8 x 10-12. )
nikitadnepr [17]

The molar solubility in water of Ag₂CrO₄ is 2.52 * 10⁻⁴ M .

<h3>How do you define Solubility ?</h3>

The measure of the degree to which a substance gets dissolved in a solvent to become a solution.

The formula for determination of molar solubity is

\rm K_{sp} =  [A^+]^a [B^-]^b

\rm K_{sp}          = solubility product constant

\rm [A^+]^a  =     cation in an aqueous solution

\rm  [B^-]^b  = anion in an aqueous solution

a, b          = relative concentrations of a and b

We have the data given as

\rm K_{sp} for Ag₂CrO₄ is 8 x 10⁻¹²

For Silver chromate , the solubility would be only one-half of the Ag⁺ concentration.

We denote the solubility of Ag₂CrO₄ as S mol L⁻¹.

Then for a saturated solution, we have

[Ag⁺]=2S

[CrO₂⁴⁻]=S

the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction

(2S)²(S)=4S³ = 8 * 10⁻¹²

S=∛ (2*10⁻¹²)

S= 1.26 * 10⁻⁴ M

2S = 2.52 * 10⁻⁴ M

Therefore the molar solubility in water of Ag₂CrO₄ is 2.52 * 10⁻⁴ M .

To know more about Solubility

brainly.com/question/8591226

#SPJ4

4 0
2 years ago
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2CO +0, 200, what is the mole ratio of carbon monoxide to carbon
Nastasia [14]

Answer:

B. 2.2

Explanation:

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3 years ago
The middle block of the periodic table is called?​
Masteriza [31]

Answer: The transition metals make up the middle block of he periodic table

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3 years ago
Round the mmHg readings to the nearest whole number. Round the kPa readings to the nearest tenth. Round the atm readings to the
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6 0
2 years ago
Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25
Veseljchak [2.6K]

Answer: The molecular formula will be C_6H_6O_6

Explanation:

Mass of CO_2 = 17.95 g

Mass of H_2O= 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, =\frac{12}{44}\times 17.95=4.89g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, =\frac{2}{18}\times 4.87=0.541g of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H =  0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.407}{0.407}=1

For H =\frac{0.541}{0.407}=1

For O = \frac{0.410}{0.407}=1

The ratio of C : H : O = 1: 1  : 1

Hence the empirical formula is CHO.

Hence the empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = \frac{44.0}{0.25}\times 1=176g

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6

The molecular formula will be=6\times CHO=C_6H_6O_6

5 0
3 years ago
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