The answer is C. statement of owner's equity
Answer:
See explanation below
Explanation:
To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:
m = m₀e^-kt (1)
In this case, k will be the constant rate of this element. This is calculated using the following expression:
k = ln2/t₁/₂ (2)
Let's calculate the value of k first:
k = ln2/2.7 = 0.2567 d⁻¹
Now, we can use the expression (1) to calculate the remaining mass:
m = 8.1 * e^(-0.2567 * 2.6)
m = 8.1 * e^(-0.6674)
m = 8.1 * 0.51303
m = 4.16 mg remaining
Answer:
79.3%
Explanation:
Percent yield = 5.33g/6.72g x 100% = 79.3%
Answer:
0.862 J/gºC
Explanation:
The following data were obtained from the question:
Mass of metal (Mₘ) = 50 g
Initial temperature of metal (Tₘ) = 100 °C
Mass of water (Mᵥᵥ) = 400 g
Initial temperature of water (Tᵥᵥ) = 20 °C
Equilibrium temperature (Tₑ) = 22 °C
Specific heat capacity of water (Cᵥᵥ) = 4.2 J/gºC
Specific heat capacity of metal (Cₘ) =?
The specific heat capacity of the metal can be obtained as follow:
Heat lost by metal = MₘCₘ(Tₘ – Tₑ)
= 50 × Cₘ × (100 – 22)
= 50 × Cₘ × 78
= 3900 × Cₘ
Heat gained by water = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
= 400 × 4.2 × (22 – 20)
= 400 × 4.2 × 2
= 3360 J
Heat lost by metal = Heat gained by water
3900 × Cₘ = 3360
Divide both side by 3900
Cₘ = 3360 / 3900
Cₘ = 0.862 J/gºC
Therefore, the specific heat capacity of the metal is 0.862 J/gºC
WHAT IS KNOWN AS HYBRIDIZATION
<h3> It is the change in the orbitals of the central atom of the molecule to form bonds with other atom if same type or another type.</h3>
The hybrid orbitals r of same level .
The Hybridization takes place in between orbitals of equal or very less energy levels to form same level of energy in all orbitals.
The orbitals combine is always equal to The number of orbitals formed
The names of hybridized orbitals r kept according to the orbital which r combined to form them
<h2 /><h2>Eg:-</h2>
sp orbital :- one s and one p combine to form sp orbital
orbital :- one s and two p orbitals combine to form it..
