The question is incomplete, the complete question is;
The table above summarizes data given to a student to evaluate the type of change that took place when substance X was mixed with water. The student claimed that the data did not provide enough evidence to determine whether a chemical or physical change took place and that additional tests were needed. Which of the following identifies the best way to gather evidence to support the type of change that occurred when water and Xwere mixed?
A. Measuring the melting point of the mixture of water and X
B. Adding another substance to the mixture of water and X to see whether a solid forms
C Measuring and comparing the masses of the water, X, and the mixture of water and X
D Measuring the electrical conductivities of X and the mixture of water and X
Answer:
D Measuring the electrical conductivities of X and the mixture of water and X
Explanation:
Unfortunately, I am unable to reproduce the table here. However, from the table, the temperature of the of the mixture of the solid X and water was 101.6°C. This is above the boiling point of water and way below the temperature of the solid X.
This goes a long way to suggest that there was some kind of interaction between the water and X which accounted for the observed temperature of the system of X in water.
The only way we can be able to confirm if X actually dissolved in water is to measure the conductivity of the water. dissolved solids increase the conductivity of water.
the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2
2135 cm3 divided by 127 degrees celcius = x divided by 206
answer: 3460 cm3
Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol.
n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
6.7g/0.28125 mol = 23.82 g/mol
The pH of a buffer solution : 4.3
<h3>Further explanation</h3>
Given
0.2 mole HCNO
0.8 mole NaCNO
1 L solution
Required
pH buffer
Solution
Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.
![\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3DKa%5Ctimes%5Cfrac%7Bmole%5C%3Aweak%5C%3Aacid%7D%7Bmole%5C%3Asalt%5Ctimes%20valence%7D)
valence according to the amount of salt anion
Input the value :
![\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3D2.10%5E%7B-4%7D%5Ctimes%5Cfrac%7B0.2%7D%7B0.8%5Ctimes%201%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D5%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5CpH%3D5-log~5%5C%5C%5C%5CpH%3D4.3)
