Answer:Its c
Explanation:
Let us use a double displacement reaction of Lead (II) Nitrate and Potassium Chromate to produce Lead (II) Chromate and Potassium Nitrate to practice balancing an equation.
We begin with the base equation provided in the question.Pb(NO3)2(aq)+K2CrO4(aq)→PbCrO4(s)+KNO3(aq)
Looking at the the atom inventory ReactantsPb=1NO3=2
K=2CrO4=1
Products
Pb=1
NO3=1
K=1
CrO4=1
We can see that the
Kand NO3
are imbalanced.
If we add a coefficient of 2 in front of the
KNO3
this will balance the equation.
Pb(NO3)2(aq)+K2CrO4(aq)→PbCrO4(s)+2KNO3(aq)
Answer:
Explanation:
An element is a _distinct____ substance represented by a __chemical____ symbol. Chemical symbols have _one__ or _two__ letters. The first letter of a chemical symbol is always a _Capital_____ letter and the second letter is always a _small_____ letter.
Elements are distint substances that cannot be split up into simpler substances. Such substances are made up of one kind of atom. Each of them is usually symbolised by a capital letter or a capital letter followed by a small letter derieved from the English or latin or greek name of the element.
POH = pKb + log [salt] / [base] ---> Basic Buffer
and pH = 14 - pOH
to predict the relative concentrations of the buffer components you need to know both pKb and pH...
While in this case only pKb is given and the pH of buffer is not given. So, nothing can be concluded about the relative concentrations of CH₃NH₂ and CH₃NH₃Cl
Answer: The percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make 350 mL of solution is 25.5%.
Explanation:
Given: Volume of solute = 120 mL
Volume of solvent = 350 mL
Now, total volume of the solution is as follows.

Let us assume that 100 mL of solution is taken and the amount of isopropyl alcohol present in it is as follows.

Hence, there is 25.53 mL isopropyl alcohol is present in 100 mL of solution. Therefore, %v/v is calculated as follows.

Thus, we can conclude that the percent by volume of isopropyl alcohol in a solution made by mixing 120 mL of the alcohol with enough water to make 350 mL of solution is 25.5%.