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3 years ago

Where are all metalloid found in this periodic table?

Chemistry

2 answers:

ale4655 [162]3 years ago

7 0

Answer:

Explanation:

THe GrEen OnEs

dimulka [17.4K]3 years ago

5 0

Answer:

They are in the green "staircase"

Explanation:

these elements are the dividers between metals (on the left of them) and non metals (on the right of them), they have some of the properties of both groups

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Which of these pairs of elements have the same number of valence electrons

erastovalidia [21]

Beryllium (Be) and Calcium (Ca)

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3 years ago

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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What is a fossil?

katrin2010 [14]

The remains of an animal or plant preserved within a layer of rock I think I’m sorry if I’m wrong

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PolarNik [594]

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How many moles of MgCl2 are there in 340 g of the compound?

iragen [17]

To calculate the amount of the compound in the units of grams, we need to first obtain for the molar mass of the compound. We calculate as follows:

MgCl2 = 24.31 + (2 x 35.45) = 95.21 g/mol

340 g MgCl2 ( 1 mol / 95.21 g ) = 3.57 mol MgCl2

Hope this answers the question. Have a nice day.

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3 years ago

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