Answer:
3 e⁻ transfer has occurred.
Explanation
This is a redox reaction.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
- [Ne]= (1s²) (2s² 2p⁶)
A combination of both the reactions( Half-reactions) leads to a redox reaction.
Let us look at initial configurations of Al and Cl
[Al]= 1s² 2s² 2p⁶ 3s² 3p¹
[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵
Hence, Al can lose 3 electrons to achieve octet config.
and, Cl can gain 1e to achieve nearest noble gas config. [Ar]
This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.
Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃
Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)
Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.
Answer:
The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
Explanation:
Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.
So, the following three equations can be written as per given information:
x+y+z = 149 ........(1)
8y+8z = 936 ........(2)
6x+7y = 535 .........(3)
From equation- (2), y+z = 
= 117
By substituting the value of (y+z) in equation- (1) we get,
x = 149-(y+z) = 149-117 = 32
By substituting the value of x into equation- (3) we get,
y = 
= 49
By substituting the value of y into equation- (2) we get,
z = (117-49) = 68
So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
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
Answer:
90g of H2O
Explanation:
2H2 + O2 —> 2H2O
First, we calculate the molar masses of H2 And H20.
Molar Mass of H2 = 2g/mol
Mass conc of H2 from the balanced equation = 2 x 2 = 4g
Molar Mass of H2O = 2 + 16 = 18g/mol
Mass conc of H2O from the balanced equation = 2x18 = 36g
From the equation,
4g of H2 produced 36g of H2O
Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O
