There are 20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.
<h3>How many molecules in 3.4 moles of NH4NO3?</h3>
We know that one mole of a substance has 6.022 × 10²³ molecules so in 3.4 moles of NH4NO3, we have 20.5 x 10^24 molecules if we multiply the 6.022 × 10²³ with 3.4.
So we can conclude that there are 20.5 x 10^24 molecules are present in 3.4 moles of NH4NO3.
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A. SO2Cl2(g) --> SO2(g) + Cl2(g)
<span>1 mole of SOCl2 becomes 1 mole SO2 and 1 mole Cl2 </span>
<span>1 mole --> 2 moles </span>
<span>entropy increases </span>
The mass of oxygen reacted/required in this reaction is obtained as 48g.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with mass- volume or mass - mole relationship which ultimately depends on the balanced reaction equation.
Now, we have the reaction; S + O2 ------>SO2
If 1 mole of sulfur dioxide contains 22.4 L
x moles of sulfur dioxide contains 33.6L
x = 1.5 moles of sulfur dioxide.
Since the reaction is 1:1, the number if moles of oxygen required/reacted is 1.5 moles.
Mass of oxygen required/reacted = 1.5 moles * 32 g/mol = 48g
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4. 2Li + 2H2O -> 2LiOH + H2
5. C6H12O6 + 6O2 -> 6CO2 + 6H2O
6. Zn + 2HCl -> ZnCl2 + H2
9. H2SO4 + Pb -> PbSO4 + H2
10. Ca(OH)2 + NH4Cl -> NH4 + CaCl2 + H2O
thats all i know
The scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.
<h3>What is a cathode ray?</h3>
A cathode ray is a tube that contains negatively charged electrode( that is the cathode) which emits electrons when heated at a low pressure.
The cathode ray was used by the scientist, J.J. Thomson to find the ratio of charge to mass (e/m) of the electrons.
Therefore, the scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.
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