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kkurt [141]
3 years ago
7

The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

90g H and 4.71g of O, is the substance water?​
Chemistry
1 answer:
Gre4nikov [31]3 years ago
6 0

Answer:

                    No the substance is not water.

Explanation:

                   The balance chemical equation for the decomposition of water is as follow;

                                           2 H₂O = 2 H₂ + O₂

Step 1: <u>Calculate moles of H₂O;</u>

               Moles  =  Mass / M.Mass

               Moles  =  5.0 g / 18.01 g/mol

               Moles  =  0.277 moles of H₂O

Step 2: <u>Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:</u>

According to equation,

                        2 moles of H₂O produced  =  1 mole of O₂

So,

                  0.277 moles of H₂O will produce  =  X moles of O₂

Solving for X,

                     X =  0.277 mol × 1 mol / 2 mol

                     X =  0.138 moles of O₂

Also,

According to equation,

                        2 moles of H₂O produced  =  2 mole of H₂

So,

                  0.277 moles of H₂O will produce  =  X moles of H₂

Solving for X,

                     X =  0.277 mol × 2 mol / 2 mol

                     X =  0.227 moles of H₂

Step 3: <u>Calculate Mass of O₂ and H₂ as;</u>

For O₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.138 mol × 31.99 g/mol

                 Mass  =  4.44 g of O₂

For H₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.227 mol × 2.01 g/mol

                 Mass  =  0.559 g of H₂

Conclusion:

                   From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

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saveliy_v [14]

Answer:- 2.40*10^1^0atoms

Solution:- It is a simple unit conversion problem. We could solve this using dimensional analysis.

We know that, 1 US dollar = 100 cents

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1g=10^6\mu g

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1penny(\frac{$1}{100pennies})(\frac{1\mu g}{$1000})(\frac{1g}{10^6\mu g})(\frac{6.02*10^2^3atoms}{251g})

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3 years ago
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Which of the following gives the definition of ion-dipole attraction? Select the correct answer below: O a. lon-dipole attractio
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6 0
3 years ago
calcualte pressure at STP in 10.0 L vessel after reaction of 1.0 L hydrochloride acid (concentration 35% and density 1.28 g/cub.
sattari [20]

Answer:

The pressure in the vessel is 13,3 atm.

Explanation:

The reaction that occurs in vessel (where limestone is 96% of CaCO₃) is:

2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)

The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.

We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:

HCl:

1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol

(Concentration)      (L to cm³)         (cm³ to g)      (g to mol)  (moles of reaction)

moles of HCl= 6,14 mol

CaCo₃:

   1,0 kg     ×       (96/100)                ×   (1000 g/1kg) × (1 mol/100,09g)

(Limestone) (CaCo₃ in limestone)          (kg to g)            (g to mol)

moles of CaCo₃= 9,59 mol

So, <em>reactant limit is HCl</em>

This reaction have a yield of 97%. So, the CO₂ moles are:

6,14 mol × 97÷ = 5,96 mol CO₂

The ideal gas formula to obtain pressure is:

P = nRT/V

Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L

Replacing this values in the equation the pressure is

P = 13,3 atm

I hope it helps!

7 0
3 years ago
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