For a neutralisation reaction...
Acid + base 》 salt + water....
The above reaction looks much alike so is a neutralisation...
It is certainly not compustion as no oxygen to react and CO2 produced
Answer:
TRUE.
Explanation:
Mass Number is the sum of protons and neutrons present in the nucleus of an atom. Isotopy is a phenonmenon that occurs when atoms of same elements have different mass number (Number of neutrons).
2H isotope has 1 proton and 1 neutron.
3H isotope has 1 proton and 2 neutrons.
This meeans 2H isotope has fewer neutrons when compared to the 3H isotope. The correct option is TRUE.
The balanced chemical reaction is:
<span>2Na + 2H2O → 2NaOH + H2
</span><span>
We first use the amount of hydrogen gas to be produced and the molar mass of the hydrogen gas to determine the amount in moles to be produced. Then, we use the relation from the reaction to relate H2 to Na.
53.2 g H2 ( 1 mol / 2.02 g ) ( 2 mol Na / 1 mol H2 ) ( 22.99 g / 1 mol ) = 1210.96 g Na
1210.96 g Na ( 1 mL / 0.97 g ) = 1248.41 mL Na needed</span>
Answer:
May I assume "ethanol acid is just ethanol (it has one slightly acidic H atom). If so, the molar mass is 46.02 g/mole.
Explanation:
We have 30 cm^3 [30 ml] of 1.0 M (1 mole/liter) [1 dm³ = 1 liter].
That is 1 mole/liter. 30 ml would contain (0.030 liter)*(1 mole/1 liter) = 0.03 moles.
Answer : The molecular weight of this compound is 891.10 g/mol
Explanation : Given,
Mass of compound = 12.70 g
Mass of ethanol = 216.5 g
Formula used :
![\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=i\times T_f\times\frac{\text{Mass of compound}\times 1000}{\text{Molar mass of compound}\times \text{Mass of ethanol}}](https://tex.z-dn.net/?f=%5CDelta%20T_f%3Di%5Ctimes%20K_f%5Ctimes%20m%5C%5C%5C%5CT_f%5Eo-T_f%3Di%5Ctimes%20T_f%5Ctimes%5Cfrac%7B%5Ctext%7BMass%20of%20compound%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20compound%7D%5Ctimes%20%5Ctext%7BMass%20of%20ethanol%7D%7D)
where,
= change in freezing point
= temperature of pure ethanol = ![-117.300^oC](https://tex.z-dn.net/?f=-117.300%5EoC)
= temperature of solution = ![-117.431^oC](https://tex.z-dn.net/?f=-117.431%5EoC)
= freezing point constant of ethanol = ![1.99^oC/m](https://tex.z-dn.net/?f=1.99%5EoC%2Fm)
i = van't hoff factor = 1 (for non-electrolyte)
m = molality
Now put all the given values in this formula, we get
![(-117.300)-(-117.431)=1\times 1.99^oC/m\times \frac{12.70g\times 1000}{\text{Molar mass of compound}\times 216.5g}](https://tex.z-dn.net/?f=%28-117.300%29-%28-117.431%29%3D1%5Ctimes%201.99%5EoC%2Fm%5Ctimes%20%5Cfrac%7B12.70g%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20compound%7D%5Ctimes%20216.5g%7D)
![\text{Molar mass of compound}=891.10g/mol](https://tex.z-dn.net/?f=%5Ctext%7BMolar%20mass%20of%20compound%7D%3D891.10g%2Fmol)
Therefore, the molecular weight of this compound is 891.10 g/mol