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Question
question 14
blank 1 grams of caco₃ is in 4 l of a 3.4 m mgoh solution
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question 15
what is the molarity of a mgcl₂ solution made by diluting 100 ml of a 1.2 m solution to a final volume of 150 ml? blank 1 m
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Question 14
Step1: Identify the problem
The amount of $CaCO_3$ in a $Mg(OH)_2$ solution is not related by a chemical reaction here. The information about the $Mg(OH)_2$ solution is extraneous. But if we assume we just want to find the amount of $CaCO_3$ and no other data is given, we cannot solve it. However, if we assume it's a mis - question and we focus on molar - mass calculations. The molar mass of $CaCO_3$ is $M = 40+12 + 3\times16=100\ g/mol$. But without moles or other relevant data, we can't calculate the mass. Since the question seems to be incomplete or has wrong setup, we assume it's asking for a non - existent relationship. If we assume we just need to state the molar mass conceptually, we note the molar mass. But if we assume a normal mass - calculation problem with no data, we can't proceed. Let's assume a standard mass calculation problem and we need to find mass from moles. Since no moles are given, we can't calculate. But if we assume we are just showing the process with no real data, we know $m=n\times M$.
Step2: Analyze the given data
We have no data related to moles of $CaCO_3$ in the solution. The volume and molarity of $Mg(OH)_2$ are not relevant to finding mass of $CaCO_3$ without a chemical reaction context which is not given.
Step1: Recall the dilution formula
The dilution formula is $M_1V_1 = M_2V_2$, where $M_1$ is the initial molarity, $V_1$ is the initial volume, $M_2$ is the final molarity and $V_2$ is the final volume.
Step2: Identify the given values
$M_1 = 1.2\ M$, $V_1=100\ mL = 0.1\ L$, $V_2 = 150\ mL=0.15\ L$.
Step3: Solve for $M_2$
From $M_1V_1 = M_2V_2$, we can express $M_2=\frac{M_1V_1}{V_2}$. Substituting the values: $M_2=\frac{1.2\ M\times0.1\ L}{0.15\ L}$.
$M_2 = 0.8\ M$
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