Answer: Water displacement
Explanation:
When you put an irregularly shaped object in a graduated cylinder, the water level rises. Subtract the original number of ml from the number of ml it was displaced, and you get its volume.
Answer:
a. Raise the pH slightly
Explanation:
We know that
Pka of HNO2/KNO2 =3.39
Moles of HNO2 in the buffer=0.247 mol/L×1L=0.247 moles
Moles of NO2-=0.329mol/L×1L=0.329 moles
If 0.271 moles of Ca(OH)2 is added it will neutralise 0.136 moles of acid ,HNO2,remaining HNO2=0.247-0.136=0.111 moles
Moles of NO2- will increase as 0.0333 moles Ca(NO)2 will be formed =0.0333+0.036=0.0693 moles
pH=pka+log [base]/[acid] {henderson -hasselbach equation}
=3.39+log (0.0693/0.0317)=3.39+0.34=3.73
pH=3.73
Answer: A.Light travels in a straight line., B.Light behaves in a predictable way. , D.Light curves around corners or obstructions
-Hope this helps<3
<u>Answer:</u> The pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.
<u>Explanation:</u>
There are three types of solution: acidic, basic and neutral
To determine the type of solution, we look at the pH values.
- The pH range of acidic solution is 0 to 6.9
- The pH range of basic solution is 7.1 to 14
- The pH of neutral solution is 7.
We are given:
Concentration of HI = 0.100 M
1 mole of HI produces 1 mole of hydrogen ions and 1 mole of iodide ions
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
![[H^+]=0.100M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.100M)
Putting values in above equation, we get:
To calculate the pOH of the solution, we use the equation:
pH + pOH = 14
Hence, the pH and pOH of the solution is 1 and 13 respectively and the solution is acidic in nature.
The reaction between calcium carbonate and hydrochloric acid can be expressed through the chemical reaction,
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
The molecular weight of calcium carbonate is 100 g/mol while that of hydrochloric acid is 36.45. The equation above depicts that 100 g of calcium carbonate can be dissolved in 72.9 g of hydrochloric acid.
x = (4 g HCl)(100 g CaCO3 / 72.9 HCl)
x = 5.49 g
Answer: 5.49 g
