Answer:
In the first combination neutralization takes place to give a salt. So, solution 'a' is neutral in nature.
In the solution 'c', both salts are resulted by the combination of weak base and strong acid. The combination of these salts suppresses the acidity.
In last combination basic nature is observed due to the presence of CN⁻ ions. Thus, the solution 'd' is basic in nature.
Out of the five given solutions, 0.0100 M in HF and 0.0100 M in KBr is most acidic. Therefore, solution 'b' is most acidic in nature.
Explanation:
Answer:
As the gas is heated, the particles will begin to move faster. ... Because the gas remains at a constant pressure and volume, the particles cannot spread out so they simply move around the container even faster.
I believe the answer is C
Answer:
K^+ and NO3^-
Explanation:
In a balanced ionic equation, we usually see the species that react to yield the main product in the reaction.
Consider the reaction;
Pb(NO3)2(aq) +2 KI(aq) -------> PbI2(s) + 2KNO3(aq)
The main product in this reaction is PbI2. Hence the balanced ionic equation is;
Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)
Notice that K^+ and NO3^- did not participate in this reaction. All ions that are part of the molecular equation but do not participate in the ionic reaction equation are called spectator ions. Hence K^+ and NO3^- are spectator ions in this reaction as can be seen clearly above.
Answer:
The empirical formula is ZnO2
Explanation:
What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?
Step 1: Data given
Suppose the compound has a mass of 100.0 grams
A compound contains:
67.1 % Zinc = 67.1 grams
100 - 67.1 = 32.9 % oxygen = 32.9 grams
Molar mass of Zinc = 65.38 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate moles of Zinc
Suppose the compound is 100 grams
Moles Zn = 67. 10 grams / 65.38 g/mol
Moles Zn = 1.026 moles
Step 3: Calculate moles of O
Moles O = 32.90 grams / 16.00 g/mol
Moles O = 2.056 moles
Step 4: Calculate mol ratio
We divide by the smallest amount of moles
Zn: 1.026/1.026 = 1
O: 2.056/1.026 = 2
The empirical formula is ZnO2
To control this we can calculate the % Zinc for 1 mol
65.38 / (65.38+2*16) = 0.67.1 = 67.2 %
