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3 years ago

Section I: Experimental Overview

Chemistry

1 answer:

Mariana [72]3 years ago

5 0

Answer:

If you constructed graphs, what trends do they indicate in your data?

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Which of the following elements has the highest electronegativity?

Nastasia [14]

Answer: is c

Explanation:

8 0

3 years ago

Read 2 more answers

How many grams of iron can be made with 21.5g of Fe2O3

SIZIF [17.4K]

The mass (in grams) of iron, Fe that can be made from 21.5 g of Fe₂O₃ is 15.04 g

We'll begin by writing the balanced equation for the reaction. This is given below:

2Fe₂O₃ -> 4Fe + 3O₂

Molar mass of Fe₂O₃ = 159.7 g/mol
Mass of Fe₂O₃ from the balanced equation = 2 × 159.7 = 319.4 g
Molar mass of Fe = 55.85 g/mol
Mass of Fe from the balanced equation = 4 × 55.85 = 223.4 g

From the balanced equation above,

319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe

<h3>How to determine the mass of iron, Fe produced</h3>

From the balanced equation above,

319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe

Therefore,

21.5 g of Fe₂O₃ will decompose to produce = (21.5 × 223.4) / 319.4 = 15.04 g of Fe

Thus, 15.04 g of Fe were produced.

Learn more about stoichiometry:

brainly.com/question/9526265

#SPJ1

5 0

1 year ago

How many moles of CCI are there in 78.2 g of CCI.?

vaieri [72.5K]

Answer:

0.508 mole

Explanation:

NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.

The number of mole present in 78.2 g of CCl₄ can be obtained as follow:

Mass of CCl₄ = 78.2 g

Molar mass of CCl₄ = 12 + (35.5×4)

= 12 + 142

= 154 g/mol

Mole of CCl₄ =?

Mole = mass / molar mass

Mole of CCl₄ = 78.2 / 154

Mole of CCl₄ = 0.508 mole

Therefore, 0.508 mole is present in 78.2 g of CCl₄

6 0

2 years ago

Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

Now put all the given values in this formula, we get:

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago

15.True or False:

Nadusha1986 [10]

True
(I hate the 20 character limit)

3 0

2 years ago