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3 years ago

9

Complete combustion of 6.4g of compound K produced 8.8 g of carbon dioxide and 7.2 g of water. Calculate the empirical formula o

f K.

1 answer:

Elenna [48]3 years ago

7 0

Lh (x1)78.009+1= 0.0000567

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Calculate the number of moles in 25.0 g of each of the

mamaluj [8]

Answer:

A. 6.25moles

B. 0.78moles

C. 0.32moles

D. 0.15moles

E. 0.43moles

Explanation:

use

n = mass/molar mass

A. He

25/4 = 6.25moles

B. O2

25/32 = 0.78moles

C. Al(OH)3

27+ 3(17) = 78
25/78 = 0.32 moles

D. GaS3

70+3(32) = 166
25/166 = 0.15moles

E. C4H10

4(12)+10(1) =58
25/58 = 0.43moles

5 0

2 years ago

Select all the words that describe what the elements have in common if they are located on the same PERIOD.

Marrrta [24]

Answer:

Orbitals

Shells

Energy Levels

Explanation:

Elements in the same period will have the same energy levels which is the same as orbitals and shells.

The period of an element indicates the energy level in an atom.
Elements in period 1 will have one energy level or shell or orbitals.
Down a group this is not the case.
The energy level increases from top to bottom.
As we go down the group, we are transiting from one period to another.

4 0

3 years ago

Please help I will reward brainly

nordsb [41]

Your answer is yeast, since it is a single celled organism that can be classified into a subkingdom related to sac fungi.

6 0

3 years ago

Why is potasium oxide not used to investigate property of hydrogen gas

Anarel [89]

Because potassium is more reactive than hydrogen

3 0

3 years ago

Under certain conditions, the equilibrium constant of the reaction below is Kc=1.7×10−3. If the reaction begins with a concentra

svetoff [14.1K]

Answer:

[Cl2] equilibrium = 0.0089 M

Explanation:

<u>Given:</u>

[SbCl5] = 0 M

[SbCl3] = [Cl2] = 0.0546 M

Kc = 1.7*10^-3

<u>To determine:</u>

The equilibrium concentration of Cl2

<u>Calculation:</u>

Set-up an ICE table for the given reaction:

$SbCl5(g)\rightleftharpoons SbCl3(g)+Cl2(g)$

I 0 0.0546 0.0546

C +x -x -x

E x (0.0546-x) (0.0546-x)

$Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M$

The equilibrium concentration of Cl2 is:

= 0.0546-x = 0.0546-0.0457 = 0.0089 M

5 0

3 years ago

Read 2 more answers