Answer:
its 2 parts creating a group which would be combustion
Answer:
The percent composition is 21% N, 6% H, 24% S and 49% O.
Explanation:
1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.
In 1 mole of (NH4)2SO4, there are:
- 2 moles of N.
- 8 moles of H.
- 1 mole of S.
- 4 moles of O.
2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:
- 2 moles of N (14g/mol) = 28g
- 8 moles of H (1g/mol) = 8g
- 1 mole of S (32g/mol) = 32g
- 4 moles of O (16g/mol) = 64g
3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:
In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.
So, the percent composition is 21% N, 6% H, 24% S and 49% O.
Answer:
NH3 is polar due to the bonds between nitrogen and hydrogen which have different electronegativity and due also to its asymmetrical shape.
Explanation:
NH3 is polar as there are 3 dipoles in the ammonia molecule that do not balance each other out.
Considering the N-H bond which is polar because N with an electronegativy of 3.0, is more electronegative than H, with an electronegativity of 2.1. The is overall asymmetrical shape of NH3
means that the dipoles remains unbalanced and do cancel out each other making the NH3 polar.
Answer:
I don't <u>understand</u><u> </u><u>your</u><u> </u><u>question</u><u> </u>
Answer:
Mg
Explanation:
The standard reduction potentials are
<u>E°/V
</u>
Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42
Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85
Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Mg2+(aq) + 2e- ⟶ Mg(s); -2.38
The more negative the standard reduction potential, the stronger the metal is as a reducing agent.
Mg is the only metal with a standard reduction potential lower than that of Cu, so
Only Mg will react spontaneously with Cu²⁺.
