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2 years ago

I pls help me this is the question:define the following terms below

Chemistry

1 answer:

Travka [436]2 years ago

8 0

Answer:

natural resources: materials or substances such as minerals, forests, water, and fertile land that occur in nature and can be used for economic gain.

renewable resources: a natural resource that is unlimited or that is naturally replenished rather quickly, such as sunlight or water

non renewable resources: (also called a finite resource) is a natural resource that cannot be readily replaced by natural means at a pace quick enough to keep up with consumption.

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A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

Answer: The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

Explanation:

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$

Hence, the concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

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Calculate the ph of the solution resulting by mixing 20.0 ml of 0.15 m hcl with 20.0 ml of 0.10 m koh

Aliun [14]

Answer:

1.60.

Explanation:

The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.

The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.

Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.

∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.

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3 years ago

I pls help me this is the question:define the following terms below​

I pls help me this is the question:define the following terms below