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Snowcat [4.5K]
3 years ago
5

The diagram shows two molecules (A and B) with different

Chemistry
1 answer:
Anarel [89]3 years ago
5 0

Answer: chromatography

Explanation:

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If 14.3 moles of H2O2 is decomposed, how many grams of oxygen gas are produced?
Studentka2010 [4]
The balanced chemical reaction would be as follows:

2H2O2 = 2H2O + O2

We are given the amount of the peroxide that decomposes. Using this as the starting point for the calculations, we can determine the amount of O2 produced. We do as follows:

14.3 mol H2O2 ( 1 mol O2 / 2 mol H2O2 ) = 7.15 mol O2 produced
5 0
2 years ago
Read 2 more answers
At this temperature, 0.300 mol H 2 0.300 mol H2 and 0.300 mol I 2 0.300 mol I2 were placed in a 1.00 L container to react. What
Alekssandra [29.7K]

Answer:

The concentration of HI present at equilibrium is 0.471 M.

Explanation:

5 0
3 years ago
Nevermind i got it dont need help thanks
OverLord2011 [107]

Answer:

Ok then...

Explanation:

4 0
2 years ago
For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.
vichka [17]

Answer:

\Delta G =-103.95kJ

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

\Delta G =\Delta H-T\Delta S

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ

Best regards!

6 0
2 years ago
13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
6 0
2 years ago
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