for each problem below, write the equation and show your work. always use units and box in your final…
for each problem below, write the equation and show your work. always use units and box in your final answer.\n1. the density of silver (ag) is 10.5 g/cm³. find the mass of ag that occupies 965 cm³ of space.\nm = d·v m = 10.5·965 = 10,132.5\n10,132.5 g\n2. a 2.75 kg sample of a substance occupies a volume of 250.0 cm³. find its density in g/cm³.\nd = m/v d = 2.75/250 = 0.011\n11 g/cm³\n3. under certain conditions, oxygen gas (o₂) has a density of 0.00134 g/ml. find the volume occupied by 250.0 g of o₂ under the same conditions.\nv = m/d v = 250/0.00134 = 186,567.2\n186,567.2 ml\n4. find the volume that 35.2 g of carbon tetrachloride (ccl₄) will occupy if it has a density of 1.60 g/ml.\nv = m/d v = 35.2/1.60 = 22\n22 ml\n5. the density of ethanol is 0.789 g/ml at 20°c. find the mass of a sample of ethanol that has a volume of 150.0 ml at this temperature.\nm = d·v m = 0.789·150.0 = 118.4\n118.4 g\n6. 30.0 g of each of the following acids are needed. find the volume of each that must be measured out in a graduated cylinder.\na. hydrochloric acid (hcl), density = 1.164 g/ml\nv = m/d v = 30.0/1.164 = 25.8\n25.8 ml\nb. sulfuric acid (h₂so₄), density = 1.834 g/ml\nv = m/d v = 30.0/1.834 = 16.4\n16.4 ml\nc. nitric acid (hno₃), density = 1.251 g/ml\nv = m/d v = 30.0/1.251 = 24\n24 ml
Answer
Explanation:
Step1: Recall density - mass - volume formula
The formula is $M = D\times V$ (where $M$ is mass, $D$ is density, and $V$ is volume) or $V=\frac{M}{D}$ when solving for volume.
Step2: Solve problem 1
Given $D = 10.5\ g/cm^{3}$ and $V = 965\ cm^{3}$, using $M = D\times V$, we have $M=10.5\times965 = 10132.5\ g$.
Step3: Solve problem 2
First convert mass to grams: $2.75\ kg=2750\ g$. Given $M = 2750\ g$ and $V = 250.0\ cm^{3}$, using $D=\frac{M}{V}$, we get $D=\frac{2750}{250}=11\ g/cm^{3}$.
Step4: Solve problem 3
Given $M = 250.0\ g$ and $D = 0.00134\ g/mL$, using $V=\frac{M}{D}$, we have $V=\frac{250}{0.00134}\approx186567.2\ mL$.
Step5: Solve problem 4
Given $M = 35.2\ g$ and $D = 1.60\ g/mL$, using $V=\frac{M}{D}$, we get $V=\frac{35.2}{1.60}=22\ mL$.
Step6: Solve problem 5
Given $D = 0.789\ g/mL$ and $V = 150.0\ mL$, using $M = D\times V$, we have $M=0.789\times150.0 = 118.35\ g\approx118.4\ g$.
Step7: Solve problem 6A
Given $M = 30.0\ g$ and $D = 1.164\ g/mL$, using $V=\frac{M}{D}$, we get $V=\frac{30.0}{1.164}\approx25.8\ mL$.
Step8: Solve problem 6B
Given $M = 30.0\ g$ and $D = 1.834\ g/mL$, using $V=\frac{M}{D}$, we have $V=\frac{30.0}{1.834}\approx16.4\ mL$.
Step9: Solve problem 6C
Given $M = 30.0\ g$ and $D = 1.251\ g/mL$, using $V=\frac{M}{D}$, we get $V=\frac{30.0}{1.251}\approx24\ mL$.
Answer:
- $10132.5\ g$
- $11\ g/cm^{3}$
- $186567.2\ mL$
- $22\ mL$
- $118.4\ g$ 6A. $25.8\ mL$ 6B. $16.4\ mL$ 6C. $24\ mL$