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lana66690 [7]
3 years ago
8

How did the different reform movements we've discussed change and shape America? Pls help

Chemistry
1 answer:
tigry1 [53]3 years ago
6 0

Answer:

ok ............bu am not so sure

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When heated, KClO3 decomposes into KCl and O2. KClO3--> 2KCl+ 3O . If this reaction produced 62.6 g of KCl, how much O2 was p
Vladimir [108]
The solution to your problem is as follows:

 <span>2 KClO3 → 2 KCl + 3 O2 

</span><span>MW of KCL = 39.1 + 35.35 = 74.6 g/mole 

62.6/74.6 = 0.839 moles KCl produced 

0.839 moles KCl x 3O2/2KCl x 32g/mole O2 = 40.27g of O2 was produced
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Therefore, there are <span>40.27g of O2 produced during the reaction.
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly.


5 0
3 years ago
A patient needs to take 625 mg of ibuprofen twice daily. The pills in the bottle are each 250. mg. How many pills does the patie
WINSTONCH [101]
2 and a half
250 x 2 =500
250/2=125
and 125 is a half of 250
so your answer is
2 and a half
5 0
3 years ago
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Will hydrogen ever make a double bond?
Andrei [34K]
No, hydrogen can only hold one bond and that's it.  It only needs to be paired with one bond.  
7 0
3 years ago
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How many grams of oxygen gas must react to give 2.10g of ZnO?
Advocard [28]
The chemical reaction is written as:

2Zn + O2 = 2ZnO

We are given the amount of the product to be produced from the reaction. We use this value and the relation of the substances in the reaction to calculate what is asked. We do as follows:

2.10 g ZnO ( 1 mol / 81.408 g ) ( 1  mol O2 / 2 mol ZnO ) ( 32 g / 1 mol ) = 0.414 g O2 is needed
8 0
3 years ago
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

Now put all the given values in this expression, we get:

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

Now put all the given values in this expression, we get:

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
3 years ago
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