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ANTONII [103]
3 years ago
5

Lewis dot structure for Ca(OH)2

Chemistry
1 answer:
denis23 [38]3 years ago
6 0
I hope this helps you

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What weight of sodium hydroxide will react with 73 gram of hydrogen chloride gas at NTP to Produce 117.0 gram of Nacl and 36 gra
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Let's write the equation

\\ \sf\longmapsto {NaOH\atop ?}+{HCl\atop 73g}\longrightarrow {NaCl\atop 117g}+{H_2O\atop 36g}

According to law of conservation of mass .

  • Mass of products=Mass of reactants

Let required value be x

\\ \sf\longmapsto x+73=117+36

\\ \sf\longmapsto x+73=153

\\ \sf\longmapsto x=153-73

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2 years ago
He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o
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Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

Solution :  

The relation between the rate constant the activation energy is,  

\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = initial rate constant = 4.55\times 10^{-5}L/mole\text{ s}

K_2 = final rate constant = 8.75\times 10^{-3}L/mole\text{ s}

T_1 = initial temperature = 195^oC=273+195=468K

T_2 = final temperature = 258^oC=273+258=531K

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

Ea=17.285\times 10^4kJ/mole

Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

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What data should be plotted to show that experimental concentration data fits a first-order reaction? g?
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