Answer:
Double replacement reaction
Explanation:
This is what is known as a double replacement reaction. The two parts of each molecule separate and recombine to form two new compounds. For instance, 
separates into 
and 
, while 
separates into 
and 
, and then they recombine with the other compound.
Hope this helps!
Isnt the basic particle that all elements are made of electrons because i know that you can get isotopes which is a whole different subject
Answer:
[H⁺] = 1.0 x 10⁻¹² M.
Explanation:
∵ [H⁺][OH⁻] = 10⁻¹⁴.
[OH⁻] = 1 x 10⁻² mol/L.
∴ [H⁺] = 10⁻¹⁴/[OH⁻] = (10⁻¹⁴)/(1 x 10⁻² mol/L) = 1.0 x 10⁻¹² M.
∵ pH = - log[H⁺] = - log(1.0 x 10⁻¹² M) = 12.0.
∴ The solution is basic, since pH id higher than 7 and also the [OH⁻] > [H⁺].
I think- IDK
Answer: 90.04°C
Explanation: <u>Calorimeter</u> is a device measures the amount of heat of a chemical or physical process. An ideal calorimeter is one that is well-insulated, i.e., prevent the transfer of heat between the calorimeter and its surroundings. So, the net heat change inside the calorimeter is zero:
Rearraging, it can be written as
showing that the heat gained by Substance 1 is equal to the energy lost by Substance 2.
In our case, water is gaining heat, because its temperature has risen and so, brass is losing energy:
Calculating:
![m_{w}.c_{w}.\Delta T=-[m_{b}.c_{b}.\Delta T]](https://tex.z-dn.net/?f=m_%7Bw%7D.c_%7Bw%7D.%5CDelta%20T%3D-%5Bm_%7Bb%7D.c_%7Bb%7D.%5CDelta%20T%5D)
![100.4.18.(18.4-15)=-[52.9.0.375.(18.4-T)]](https://tex.z-dn.net/?f=100.4.18.%2818.4-15%29%3D-%5B52.9.0.375.%2818.4-T%29%5D)
Note: final temperature is the same as the substances are in thermal equilibrium.
Solving:
418(3.4)= - 365.01 + 19.8375T
19.8375T = 1786.21
T = 90.04
The initial temperature for the sample of brass was 90.04°.
Answer:
a. 50KCal
b. 400KCal
c. Same as (a) above
Explanation:
Given
To raise the temperature of 1kg of liquid water at 1°C requires 1KCal
To raise the temperature of 1kg of ice or water vapour by 1°C requires 0.5KCal
To melt 1kg of ice at 0°C requires 80KCal
To evaporate 1kg of liquid water sitting at 100°C requires 540KCal
a. How much heat is required to raise the temperature of 5 kg of liquid water by 20 C?
To raise the temperature of 5 kg of ice by 20°C requires:
5 kg * (0.5 kcal / kgC) * 20C
= 50 KCal
b. How much heat is required to melt 5 kg of ice at 0 C?
To melt an ice of 5 kg of ice at 0 C requires:
5 kg * (80 kcal / kg)
= 400 KCal
c. Same as (a) above
