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3 years ago

why is it easier for water to flow through rock and sediment with larger pores than through rock and sediment with smaller pores

Chemistry

2 answers:

andrey2020 [161]3 years ago

6 0

Answer:

Because of its permeability.

Explanation:

It is on quizlet. I don't think it is right though, I am sorry.

joja [24]3 years ago

6 0

Answer:

Larger pores mean less friction between flowing water and the sides of the pores. Smaller pores mean more friction between flowing water and pore walls and more twists and turns for the water to navigate. A permeable material has a greater number of larger, well-connected pore spaces. An impermeable material has fewer, smaller pores that are poorly connected.

Explanation:

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What is the total number of moles represented by 20 grams of CACO3

Nutka1998 [239]

Answer:

B. 0.2.

Explanation:

We can use the relation:

n = mass/molar mass

mass of CaCO₃ = 20 g, molar mass of CaCO₃ = 100.0869 g/mol.

∴ n = mass/molar mass = (20 g)/(100.0869 g/mol) = 0.1998 ≅ 0.2 mol.

So, the right choice is: B. 0.2.

8 0

3 years ago

Which of the following statements about trends in solubility is accurate?

Kipish [7]

"The solubility of gases decreases as temperature rises" statements about trends in solubility is accurate.

Option: D

Explanation:

A substance's solubility is the quantity of that component that is needed at a defined degree of temperature to produce a saturated solution in any set quantity of solvent. Some compounds like hydrochloric acid, ammonia, etc have solubility that reduces with rising temperature. They are both standard-pressure gases.

When heating a solvent with a gas absorbed in it, both the solvent and the solute spike in the kinetic energy.When the gaseous solute's kinetic energy rises, the molecules have a higher propensity to overcome the solvent molecules' connection and migrate to the gas phase. Thus, a gas's solubility reduces with rising temperature.

4 0

3 years ago

Calculate the number of moles of caco3 (calcium carbonate, or limestone) in a 20.0g sample of this substance

Brrunno [24]

First calculate for the molar mass of the given formula unit, CaCO₃. This can be done by adding up the product when the number of atom is multiplied to its individual molar mass as shown below.

molar mass of CaCO₃ = (1 mol Ca)(40 g Ca/mol Ca) + (1 mol C)(12 g of C/1 mol of C) + (3 mols of O)(16 g O/1 mol O) = 100 g/mol of CaCO₃

Then, divide the given amount of substance by the calculated molar mass.
number of moles = (20 g)(1 mol of CaCO₃/100 g)
number of moles = 0.2 moles of CaCO₃

Answer: 0.2 moles

6 0

3 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.