Answer:
15.438g H2O
Explanation:
First you need to find the reaction equation:
2H2O+2Na=2NaOH + H2
Hydrogen is a diatomic molecule so it will have a subscript of 2 on the right hand side. From there we can balance the reaction.
Then we can use stoichiometry:
34.2g NaOH * (1 mol NaOH/39.908g NaOH) * (2 mol H2O/2 mol NaOH) * (18.015g H2O/1 mol H20) = 15.438g H2O
It is important that when you use stoichiometry that all your units cancel out until you only have the unit you want.
Au^2S^3+ 3H^2 = 2Au + 3H^2S
The rate constant is mathematically given as
K2=2.67sec^{-1}
<h3>What is the Arrhenius equation?</h3>
The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

Therefore
KT1= 0.0110^{-1}
T1= 21+273.15
T1= 294.15K
T2= 200
T2=200+273.15
T2= 473.15K
Ea= 35.5 Kj/Mol
Hence, in j/mol R Ea is
Ea=35.5*1000 j/mol R

K2/0.0110 =e^(5.492)
K2/0.0110 =242.74
K2= 242.74*0.0110
K2=2.67sec^{-1}
In conclusion, rate constant
K2=2.67sec^{-1}
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Air is a mixture, Its constituens can be seperated
<u>Answer:</u> The solubility of oxygen at 682 torr is 
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:

Or,

where,
are the initial concentration and partial pressure of oxygen gas
are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr

Putting values in above equation, we get:

Hence, the solubility of oxygen gas at 628 torr is 