Answer:
durage
Explanation:
durage just doesn't make sense compared to the other 3
Answer:
1. MgCl2 + Zn -> Mg + ZnCl2
2. 2Al + 6H2O(g) -> 2Al(OH)3 + 3H2
3. 2Cd + O2 -> 2CdO
4. I2 + KF -> it's not going to react
5. Zn + H2SO4 -> ZnSO4 + H2
6.2KBr + Cl2 -> 2KCl + Br2
7. AgNO3 + Na -> NaNO3 + Ag
8. 2NaCl + F2 -> 2NaF + Cl2
9. AgNO3 + Mg(NO3)2 + Ag
10. Ni + H2SO4 -> NiSO4 + H2
11. Al + K2SO4
12. FeCl3 + 3Mn-> 3MnCl + Fe
13. 2Na + 2H2O -> 2NaOH + H2
14. 2K + MgBr2 -> 2KBr + Mg
15. Zn + Pb(NO3)2 -> Pb + Zn(NO3)2
16. 2AlBr3+ 3Cl2 -> 2AlCl3 + 3Br2
I'm just doing the ones that you don't have numbers already for.
2.) just leave it alone and it's correct
3.) Mg + 2AgNo3 --> Mg(No3)2 + 2Ag
5.) just leave it alone and it's correct
8.) 10C3H8O3 + 15O2 --> 30CO2 + 4H2O
10.) P4 + 6Br2 --> 4PBr3
12.) 2FeCl3 + 6NaOH --> 2Fe(OH)3 + 6NaCl
13.) 2CH3OH + 3O2 --> 2CO2 + 4H2O
14.) 2Al + 3Cu(NO3)2 --> 2Al(NO3)3 + 3Cu
15.) 3CaCl2 + 2K3AsO4 --> Ca3(AsO4)2 + 6KCl
16.) 2NH3 --> N2 + 3H2
17.) 2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
19.) Mg3N2 + 6H2O --> 3Mg(OH)2 + 2NH3
I hope this helps you!!
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
Answer:5.309 × 10²⁴ atoms.
Explanation:
Given that
molar mass of NH3 = 17
g/mol
Mass of NH3 = 5g
Therefore, No of moles of NH3 = Mass/ molar mass
= 5g/ 17g/mol
= 0.294 moles.
I mole = 6.02 × 10²³ atoms
Therefore the number of hydrogen atoms in a 0.294 moles of ammonia gives us
0.294× 6.02 × 10²³ × 3 ( since there are 3 hydrogens in Ammonia )
= 5.309 × 10²⁴ atoms.
