Answer: The ratio of the number of oxygen molecules to the number of nitrogen molecules in these flasks is 1: 1
Explanation:
According to avogadro's law, equal volumes of all gases at same temperature and pressure have equal number of moles.
According to avogadro's law, 1 mole of every substance contains avogadro's number of particles.
Thus as oxygen and nitrogen are at same temperature and pressure and are in equal volume flasks , they have same number of moles and thus have same number of molecules.
The ratio of the number of oxygen molecules to the number of nitrogen molecules in these flasks is 1: 1
It would probably stop moving. Earth has motion and we do do. without the world moving, there would be No wind at all.
Answer:
copper
Explanation:
so for this you can work out the mass for both and compare
so mass = moles × mr
so mass of sodium = 1 × 23= 23 g
and mass of copper = 1 × 63.5= 63.5 g
so copper have more mass :)
Answer:
1.8 moles of NaCl must be produced.
Explanation:
Based on the reaction:
HCl + NaOH → NaCl + H2O
<em>1 mol of HCl reacts with 1 mol of NaOH to produce 1mol of NaCl</em>
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To solve this question we must find, as first, the <em>limiting reactant:</em>
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1.8 moles of HCl will need 1.8 moles of NaOH for a complete reaction (Ratio of reaction 1:1). As there are 3.3 moles of NaOH,
<em>HCl is limiting reactant</em>
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When the 1.8 moles of HCl react completely,
1.8 moles of NaCl must be produced because 1 mole of HCl produce 1 mole of NaCl
Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂: