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2 years ago

An element’s properties can be predicted from _____.

Chemistry

2 answers:

Mrrafil [7]2 years ago

7 0

Third option is the answer

jek_recluse [69]2 years ago

5 0

It can be predicted from the location on the periodic table.

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Which is an element? ... air ...carbon dioxide ...hydrogen ...water Description

Natalka [10]

All Elements are obtained in the periodic table & water is Not an element it is a compound.

5 0

2 years ago

Read 2 more answers

How many moles do you have in 37.3 g of Co(CrO4)3

shutvik [7]

Answer:

0.0917 mol Co(CrO₄)₃

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

37.3 g Co(CrO₄)₃

Step 2: Identify Conversions

Molar Mass of Co - 58.93 g/mol

Molar Mass of Cr - 52.00 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol

Step 3: Convert

$37.3 \ g \ Co(CrO_4)_3(\frac{1 \ mol \ Co(CrO_4)_3}{406.93 \ g \ Co(CrO_4)_3} )$ = 0.091662 mol Co(CrO₄)₃

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃

8 0

2 years ago

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = $1.83g/cm^3=1.83g/ml$

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml$

Now we have to calculate the molarity of solution.

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L$

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg$