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IgorC [24]
2 years ago
12

How many gram of na2so4 are produced from 10.0 grams of NaOH

Chemistry
1 answer:
o-na [289]2 years ago
6 0

The mass of the product obtained is 17.8 g.

<h3>What is the reaction?</h3>

We know that the reaction between sodium hydroxide and sulfuric acid is given by;  2NaOH + H_{2} SO4------- > Na_{2} SO4 + H_{2}O. This is an example of a neutralization reaction in which salt and water only is produced.

Mass of the NaOH = 10g

Number of moles of the NaOH = 10g/40 g/mol = 0.25 moles

2 moles of NaOH produced 1 mole of Na2SO4

0.25 moles of NaOH produced 0.25 moles *  1 mole/2 moles

= 0.125 moles

Mass of the Na2SO4 produced =  0.125 moles * 142 g/mol

= 17.8 g

Learn more about reaction:brainly.com/question/12084405

#SPJ1

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Two common methods to generate an aldehyde is by oxidation of an alcohol and through ozonolysis.
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Answer:

<u>a</u><u>.</u><u> </u><u>True</u><u>.</u>

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of <u>e</u><u>t</u><u>h</u><u>a</u><u>n</u><u>o</u><u>l</u><u>:</u>

<u>{ \bf{CH _{3} CH_{2}OH \:  \:  \frac{Ag/O_{2} }{500 \degree C}  >  \:  \:CH _{3} CHO}}</u>

<u>{ \sf{CH _{3} CHO \:  \: is \: ethanal}}</u>

<u>B</u><u>y</u><u> </u><u>o</u><u>z</u><u>o</u><u>n</u><u>o</u><u>l</u><u>y</u><u>s</u><u>i</u><u>s</u><u>:</u>

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

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5 0
3 years ago
Read 2 more answers
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4 0
3 years ago
The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C
avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

q=c\times (T_{final}-T_{initial})

where,

q = heat gained = ?

c = specific heat = 5.20kJ/^oC

T_{final} = final temperature = 27.43^oC

T_{initial} = initial temperature = 22.93^oC

Now put all the given values in the above formula, we get:

q=5.20kJ/^oC\times (27.43-22.93)^oC

q=23.4kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = \frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole

\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -2805.8kJ/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 27.43^oC=273+27.43=300.43K

Now put all the given values in the above formula, we get:

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K

\Delta U=-2805.8kJ/mol-0

\Delta U=-2805.8kJ/mol

Therefore, the internal energy change is -2805.8 kJ/mol

5 0
3 years ago
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