Question

science, gr. 9 atoms & elements unit - density density calculations worksheet i density = mass / volume units of density solids (g/cm³) liquids (g/ml) 1. find the unknown quantity: a) d = 3 g/ml v = 100 ml m =? d×v m = 3g/ml×100ml m = 300 b) d =? v = 950 ml m = 95 g d = m/v d = 95g/950ml d = 0.1g/ml c) d = 0.5 g/cm³ v =? m = 20 g v = m/d v = 20g/0.5g/cm³ v = 40 cm³ 2. find the unknown quantity (convert first to g or ml) a) d = 24 g/ml v = 1.2 l = 1200 ml m =? m = 28800 g m = 2.88×10⁴g b) d =? v = 100 ml m = 1.5 kg = g d = m/v d = 1500/100 d = 15g/ml c) d = v = 0.5 l = 500 ml m = 500 mg = d = m/v word problems 1. a block of aluminum occupies a volume of 15.0 ml and weighs 45.75 g. what is its

Explanation:

Step1: Recall density formula

The density formula is $d=\frac{M}{V}$, where $d$ is density, $M$ is mass and $V$ is volume. We can re - arrange it to find unknowns: $M = d\times V$ and $V=\frac{M}{d}$.

Step2: Solve 1a

Given $d = 3\ g/mL$ and $V = 100\ mL$. Using $M=d\times V$, we have $M=3\ g/mL\times100\ mL = 300\ g$.

Step3: Solve 1b

Given $V = 950\ mL$ and $M = 95\ g$. Using $d=\frac{M}{V}$, we get $d=\frac{95\ g}{950\ mL}=0.1\ g/mL$.

Step4: Solve 1c

Given $d = 0.5\ g/cm^{3}$ and $M = 20\ g$. Using $V=\frac{M}{d}$, we have $V=\frac{20\ g}{0.5\ g/cm^{3}} = 40\ cm^{3}$.

Step5: Solve 2a

First convert $V = 1.2\ L$ to $1200\ mL$. Given $d = 24\ g/mL$. Using $M=d\times V$, we get $M=24\ g/mL\times1200\ mL=28800\ g$.

Step6: Solve 2b

First convert $M = 1.5\ kg$ to $1500\ g$ and $V = 100\ mL$. Using $d=\frac{M}{V}$, we have $d=\frac{1500\ g}{100\ mL}=15\ g/mL$.

Step7: Solve 2c

First convert $V = 0.5\ L$ to $500\ mL$ and $M = 500\ mg$ to $0.5\ g$. Using $d=\frac{M}{V}$, we get $d=\frac{0.5\ g}{500\ mL}=0.001\ g/mL$.

Step8: Solve word - problem 1

Given $V = 15.0\ mL$ and $M = 45.75\ g$. Using $d=\frac{M}{V}$, we have $d=\frac{45.75\ g}{15.0\ mL}=3.05\ g/mL$.

Answer:

1a: $M = 300\ g$
1b: $d = 0.1\ g/mL$
1c: $V = 40\ cm^{3}$
2a: $M = 28800\ g$
2b: $d = 15\ g/mL$
2c: $d = 0.001\ g/mL$
Word - problem 1: $d = 3.05\ g/mL$