Br₂ (l) + 2 NaI (s) → 2 NaBr (s) + I₂ (s)
Explanation:
Reacting bromide (Br₂) with sodium iodine (NaI) will produce sodium bromide (NaBr) and iodine (I₂).
To balance the equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.
Br₂ (l) + 2 NaI (s) → 2 NaBr (s) + I₂ (s)
where:
l - liquid
s - solid
This is a single replacement reaction because an element in a compound is replaced by another element. Generally a single replacement reaction is represented as: A + BC → AC + B
Learn more about:
types of chemical reactions
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balancing chemical equations
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Answer: definite proportions.
Explanation:
1) The definite proportions law states that compounds will always have the same kind of atoms (elements) in the same mass proportion (ratios).
2) For example, a molecule of water will alwys have the same mass ratio of hydrogen atoms to oxygen atoms. That is what permits to obtain the chemical formula of the water molecule as H₂O.
The mass of the two hydrogen atoms will be in a fixed ratio respect to the mass of the oxygen atoms.
Then, if you have one reactant in less proportion than the other, respect to the ratio stated by the chemical formula of water, the former will react completely (it is the limiting reactant) with the corresponding (proportional) mass of the later. Then there will be an excess of the later reactant which will not react (will remain unchanged).
The reactants can only react in the proportion defined by the chemical formulas of the final products.
Answer:
a) ΔGrxn = 6.7 kJ/mol
b) K = 0.066
c) PO2 = 0.16 atm
Explanation:
a) The reaction is:
M₂O₃ = 2M + 3/2O₂
The expression for Gibbs energy is:
ΔGrxn = ∑Gproducts - ∑Greactants
Where
M₂O₃ = -6.7 kJ/mol
M = 0
O₂ = 0
b) To calculate the constant we have the following expression:
Where
ΔGrxn = 6.7 kJ/mol = 6700 J/mol
T = 298 K
R = 8.314 J/mol K
c) The equilibrium pressure of O₂ over M is:
Answer:
0.0845 M
Explanation:
First we <u>convert 4.27 grams of potassium iodide into moles</u>, using its <em>molar mass</em>:
- Molar Mass of KI = 166 g/mol
- 4.27 g ÷ 166 g/mol = 0.0257 mol
Now we <u>calculate the molarity of the solution</u>, using <em>the number of moles and the given volume</em>:
- Molarity = moles / liters
- Molarity = 0.0257 mol / 0.304 L = 0.0845 M
Answer:
6.88 mg
Explanation:
Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄
The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.
175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P
Step 2: Calculate the rate constant for the decay of ³²P
The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.
k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹
Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days
For first-order kinetics, we will use the following expression.
ln P = ln P₀ - k × t
ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d
P = 6.88 mg
