All categories

3 years ago

Can we break an element down anymore? A. Yes B. No

Chemistry

2 answers:

padilas [110]3 years ago

7 0

According To Question It's

B) No

<h2>Explantion ⚡</h2>

Elements cannot be broken down into a simpler substance. Likewise, one element cannot be chemically converted into a different element. Chemical elements are the simplest of substances. ...

<h2>Further More ✏️</h2>

Most elements do not exist as such in nature.

This Information Is Brought You By Some Website, Hope It Will Help You.

Thanku For The Question. (◍•ᴗ•◍)

weqwewe [10]3 years ago

5 0

Answer:

Explanation:

An element is the base of basically everything

You might be interested in

True or false: in the water cycle, water returns to Earth as condesation, usually in the form of rain or snow.

ch4aika [34]

False. Condensation forms the clouds, the way water returns to the earth is through precipitation. Hope this helps.

3 0

3 years ago

Read 2 more answers

Which one is a physical change?

lozanna [386]

Answer:

dry ice !

Explanation:

6 0

4 years ago

Read 2 more answers

In an experiment 10 grams of a substance has a volume of 24 cm. What is the

egoroff_w [7]

Explanation:

density= mass/volume

= 10/24

= 0.41 g/cm³

8 0

3 years ago

a scientist determined that an unknown element didn't react with anything. The element most likely belongs to:

Triss [41]

I believe that the answer is Noble Gases. I hope this helps. :)

7 0

3 years ago

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago