<h3>
Answer:</h3>
1026.05 kJ
<h3>
Explanation:</h3>
- The Quantity of heat is calculated by multiplying the mass of a substance by specific heat capacity and the change in temperature.
- However, condensation takes place at a constant temperature.
- Therefore, Quantity of heat is calculated by multiplying mass by the latent heat of vaporization.
In this case we are given;
Mass of steam as 454 g
To calculate the amount of heat released we must know the value of latent heat of vaporization;
The latent heat of vaporization of steam is 2260 J/g
Therefore;
Heat released, Q = 454 g × 2260 J/g
=1026040 Joules
But, 1000 joules = 1 kilo-joules
Thus, Q = 1026.04 kJ
Hence, the amount of heat released is 1026.05 kJ
Answer: The movement of heat from a warmer object to a cooler one is called heat transfer
Explanation:
Unit' is the word used to describe how something is measured. When researching a question in science, we collect data, interpret it, and share the results with other scientists. A common measurement system allows us to make direct comparisons instead of having to know things like how much a certain animal weighs.
Answer: A. ![[H_3O^+]=0.64\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D0.64%5Ctimes%2010%5E%7B-7%7DM)
B. ![[OH^-]=0.11\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.11%5Ctimes%2010%5E%7B-5%7DM)
C. ![[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B0.000775%7D%3D1.3%5Ctimes%2010%5E%7B-11%7DM)
Thus solution B is basic in nature.
Explanation:
pH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
pH for acidic solutions is less than 7, for basic solutions it is more than 7 and for neutral solutions it is equal to 7.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
or ![[H^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
A. ![[OH^-]=1.55\times 10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.55%5Ctimes%2010%5E%7B-7%7DM)
![[H_3O^+]=\frac{10^{-14}}{1.55\times 10^{-7}}=0.64\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B1.55%5Ctimes%2010%5E%7B-7%7D%7D%3D0.64%5Ctimes%2010%5E%7B-7%7DM)
![pH=-log[H_3O^+]=-log[0.64\times 10^{-7}]=7](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D%3D-log%5B0.64%5Ctimes%2010%5E%7B-7%7D%5D%3D7)
B. ![[H_3O^+]=9.43\times 10^{-9}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D9.43%5Ctimes%2010%5E%7B-9%7DM)
![[OH^-]=\frac{10^{-14}}{9.43\times 10^{-9}}=0.11\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B9.43%5Ctimes%2010%5E%7B-9%7D%7D%3D0.11%5Ctimes%2010%5E%7B-5%7DM)
![pH=-log[H_3O^+]=-log[9.43\times 10^{-9}]=8](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D%3D-log%5B9.43%5Ctimes%2010%5E%7B-9%7D%5D%3D8)
C. ![[H_3O^+]=0.000775M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D0.000775M)
![pH=-log[H_3O^+]=-log[0.000775]=3](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D%3D-log%5B0.000775%5D%3D3)
Thus solution B is basic.
