Question

at 20°c the density of mercury is 13.6 g/cm³. what is the mass of 50.0 ml of mercury at 20°c?
a 3.68 g
b 6.80×10² g
c 0.272 g
d 1.00 g/ml
e none of the above
question 4
how many cups are in a 64 - oz. pitcher of lemonade? (8 fluid oz. = 1 cup)
a 5.1×10² cups
b 56 cups
c 72 cups
d 8.0 cups
e 0.13 cups

Explanation:

Step1: Convert units

We know that \(1\ \text{mL}=1\ \text{cm}^3\). So, the volume \(V = 50.0\ \text{mL}=50.0\ \text{cm}^3\).
The formula for density is \(
ho=\frac{m}{V}\), where \(
ho\) is density, \(m\) is mass, and \(V\) is volume.
We can rearrange the formula to solve for mass: \(m=
ho\times V\).
Given \(
ho = 13.6\ \text{g/cm}^3\) and \(V = 50.0\ \text{cm}^3\).

Step2: Calculate mass

Substitute the values into the formula: \(m=13.6\ \text{g/cm}^3\times50.0\ \text{cm}^3\).
\(m = 680\ \text{g}=6.80\times 10^{2}\ \text{g}\).

For the second question:

Step1: Use unit conversion

We are given that \(8\ \text{fluid oz.}=1\ \text{cup}\).
Let \(x\) be the number of cups in \(64\ \text{oz}\).
Using the proportion \(\frac{1\ \text{cup}}{8\ \text{oz}}=\frac{x}{64\ \text{oz}}\).
Cross - multiply: \(8x=64\).

Step2: Solve for \(x\)

\(x=\frac{64}{8}=8.0\) cups.

Answer:

For the first question: B. \(6.80\times 10^{2}\ \text{g}\)
For the second question: D. \(8.0\) cups