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
The stronger the pull of gravity the greater the mass
Answer:
pKa of the histidine = 9.67
Explanation:
The relation between standard Gibbs energy and equilibrium constant is shown below as:
R is Gas constant having value = 0.008314 kJ / K mol
Given temperature, T = 293 K
Given,
So, Applying in the equation as:-
Thus,
![\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHis%5D%7D%7B%5BHis%2B%5D%7D%3De%5E%7B%5Cfrac%7B15%7D%7B-0.008314%5Ctimes%20293%7D)
![\frac{[His]}{[His+]}=0.00211](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHis%5D%7D%7B%5BHis%2B%5D%7D%3D0.00211)
Also, considering:-
![pH=pKa+log\frac{[His]}{[His+]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5BHis%5D%7D%7B%5BHis%2B%5D%7D)
Given that:- pH = 7.0
So, 
<u>pKa of the histidine = 9.67</u>
Answer:
About 547 grams.
Explanation:
We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.
To do so, we can use the initial value and convert it to grams using the molar mass.
Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

Dimensional Analysis:

In conclusion, about 547 grams of copper (II) bicarbonate is produced.