Explanation:
The given equation:
H₂ + O₂ → H₂O
A chemical equation is made up of reactants and products.
Substances that combines or decomposes to give other substances are called the reactants.
In any chemical equation, the reactants are located on the left hand side of the equation.
H₂ and O₂ are the reactants in the above equation
The results of the combination or decomposition of the reactants are the products:
H₂O is the product
→ shows the direction of the reaction
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Answer is: the compound is B₂O₃.
ω(O) = 68.94% ÷ 100%.
ω(O) = 0.6894; percentage of oxygen in the compound.
ω(X) = 31.06% ÷ 100%.
ω(X) = 0.3106; percentage of unknown element in the compound.
If we take 69.7 grams of the compound:
M(compound) = 69.7 g/mol.
n(compound) = 69.7 g ÷ 69.7 g/mol.
n(compound) = 1 mol.
n(O) = (69.7 g · 0.6894) ÷ 16 g/mol.
n(O) = 3 mol.
M(compound) = n(O) · M(O) + n(X) · M(X).
n(X) = 1 mol ⇒ M(X) = 21.7 g/mol; there is no element with this molecular weight.
n(X) = 2 mol ⇒ M(X) = 10.85 g/mol; this element is boron (B).
Answer:
1. Na + O2 → Na2O (Balanced)
2. 4Al + 3O2 → 2(Al2O3) (Balanced)
3. H2 + i2 → 2HI (Balanced)
4. Mg + 2H2O → Mg(OH)2+ H2 (Balanced)
5. 2Ca +O2 → 2CaO (Balanced)
The statement which is true about the reactivity of element with 1S²2S²2P⁶3S¹ is
it is reactive because it has to lose one electron to have a full outermost energy level.
<u><em>Explanation</em></u>
- <u><em> </em></u>Element with 1S²2S²2P⁶3S¹ electron configuration is a sodium metal.
- sodium has one electron in the outermost energy level.
- for sodium to have a full outermost energy level ( 8 electrons) it loses the 1 electron in 3S¹ to form a positively charged ion. (Na⁺)