Al + Br2 → AlBr3 To balance the Br’s, make 3 Br2 and 2 AlBr3. Al + 3 Br2 → 2 AlBr3
To balance the equation, make 2 Al.
2 Al + 3 Br2 → 2 AlBr3 ...... if you would like more information please let me know and I can help you out
The nuclear equation :
₈₂²¹⁴Pb ⇒ ₈₃²¹⁴Bi + ₋₁⁰e
<h3>Further explanation
</h3>
Given
₈₂²¹⁴Pb
beta β ₋₁e⁰ particles
Required
Nuclear equation
Solution
Radioactivity is the process of unstable isotopes to stable isotopes by decay, by emitting certain particles,
-
alpha α particles ₂He⁴
- beta β ₋₁e⁰ particles
- gamma particles ₀γ⁰
- positron particles ₁e⁰
- neutron ₀n¹
The principle used is the sum of the atomic number and mass number before and after the decay reaction is the same
The reaction
₈₂²¹⁴Pb ⇒ X + ₋₁⁰e
The element X has
-the atomic number = 82 + 1 = 83
-the mass number = 214
In the periodic system, the element with atomic number 83=Bismuth
Answer:
Primer postulado:
Así Bohr asumió que el átomo de hidrógeno puede existir solo en ciertos estados discretos, los cuales son denominados estados estacionarios del átomo. En el átomo no hay emisión de radiación electromagnética mientras el electrón no cambia de órbita.
Explanation:
Balance Chemical equation is,
P₄ + 10 Cl₂ → 4 PCl₅
According to Balance Equation,
123.89 g (1 mole) of P₄ reacts to produce = 4 Moles of PCl₅
Then,
23 g of P₄ will react with excess Cl₂ to produce = X moles of PCl₅
Solving for X,
X = (4 mol × 23 g) ÷ 123.89 g
X = 0.742 g of PCl₅
Result:
When 23 g of P₄ is reacted with excess Cl₂, 0.742 g of PCl₅ is produced.
Answer:
1.195 M.
Explanation:
- We can calculate the concentration of the stock solution using the relation:
<em>M = (10Pd)/(molar mass).</em>
Where, M is the molarity of H₂SO₄.
P is the percent of H₂SO₄ (P = 40%).
d is the density of H₂SO₄ (d = 1.17 g/mL).
molar mass of H₂SO₄ = 98 g/mol.
∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.
- We have the role that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.
<em>∴ (MV) before dilution = (MV) after dilution</em>
M before dilution = 4.78 M, V before dilution = 250 mL.
M after dilution = ??? M, V after dilution = 1.0 L = 1000 mL.
∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.
