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Andreyy89
3 years ago
9

HELP PLEASE I WILL GIVE YOU BRAINLIEST

Chemistry
1 answer:
Ksivusya [100]3 years ago
6 0
The first and Third graph
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What is the mass of 4.5 x 10^22 molecules of hydrogen peroxide (H2O2)? Show your work in the space below.
valentinak56 [21]

Given :

Number of molecules of hydrogen peroxide, N = 4.5 × 10²².

To Find :

The mass of given molecules of hydrogen peroxide.

Solution :

We know, 1 mole of every compound contains Nₐ = 6.022 × 10²³ molecules.

So, number of moles of hydrogen peroxide is :

n = \dfrac{N}{N_a}\\\\n = \dfrac{4.5\times 10^{22}}{6.022\times 10^{23}}\\\\n = 0.0747 \ moles

Now, mass of hydrogen peroxide is given as :

m = n × M.M

m = 0.0747 × 34 grams

m = 2.54 grams

Hence, this is the required solution.

6 0
3 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 20.4 g g of carbon were burned in the presence
Aleonysh [2.5K]

Answer:

C + O2 -> CO2

C 20.4/12 = 1.7

O 70.4-16.0/16=3.4

CO2 1.7 mol * 44 = 74.8

3 0
3 years ago
Valerie wants to examine the properties of a thick liquid but is finding it difficult to pour the liquid into a beaker. she need
AlladinOne [14]
B applying heat to the liquid
5 0
3 years ago
A certain polyatomic ion contains 49 protons and 50 electrons. What's the net charge of this ion?
timurjin [86]

Explanation:

B more negative charges than positive charges

6 0
2 years ago
!!!Need answer for chem homework ASAP PLS !!!!
stealth61 [152]

Answer:

The answer to your question is: kc = 6.48

Explanation:

Data

             Given                 Molecular weight

CaO =    44.6 g                    56 g

CO₂ =     26 g                       44 g

CaCO₃ = 42.3 g                  100 g

Find moles

        CaO                 56 g ----------------  1 mol

                                 44.6 g --------------   x

                               x = (44.6 x 1) / 56 = 0.8 mol

        CO₂                 44 g -----------------  1 mol

                                26 g ----------------    x

                               x = (26 x 1 ) / 44 = 0.6  moles

        CaCO₃            100 g ---------------   1 mol

                                42.3g --------------    x

                              x = (42.3 x 1) / 100 = 0.423 moles

       

Concentrations

 

        CaO     =    0.8 / 6.5  = 0.12 M

         CO₂    =     0.6 / 6.5 = 0.09 M

        CaCO₃  =   0.423 / 6.5 = 0.07 M

Equilibrium constant =   \frac{[products]}{[reactants]}

Kc = [0.07] / [[0.12][0.09]

Kc = 0.07 / 0.0108

kc = 6.48

7 0
3 years ago
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