chemistry: density problems\nfor each problem below, write the equation and show your work. always use units…

chemistry: density problems\nfor 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.\n2 a 2.75 kg sample of a substance occupies a volume of 250.0 cm³. find its density in 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.\n4 find the volume that 36.2 g of carbon tetrachloride (ccl₄) will occupy if it has a density of 1.60 g/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.\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 = 1.184 g/ml\nb sulfuric acid (h₂so₄), density = 1.834 g/ml\nc nitric acid (hno₃), density = 1.251 g/ml

chemistry: density problems\nfor 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.\n2 a 2.75 kg sample of a substance occupies a volume of 250.0 cm³. find its density in 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.\n4 find the volume that 36.2 g of carbon tetrachloride (ccl₄) will occupy if it has a density of 1.60 g/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.\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 = 1.184 g/ml\nb sulfuric acid (h₂so₄), density = 1.834 g/ml\nc nitric acid (hno₃), density = 1.251 g/ml

Answer

Explanation:

Step1: Recall density formula

The density formula is $\rho=\frac{m}{V}$, where $\rho$ is density, $m$ is mass and $V$ is volume. We can re - arrange it to find mass $m = \rho V$. Given $\rho = 10.5\ g/cm^{3}$ and $V=965\ cm^{3}$. $m=\rho V=10.5\ g/cm^{3}\times965\ cm^{3}=10132.5\ g$

Answer:

$10132.5\ g$

Explanation:

Step1: Convert mass to grams

First, convert the mass from kg to g. Since $1\ kg = 1000\ g$, a $2.76\ kg$ sample has a mass $m = 2.76\times1000\ g=2760\ g$.

Step2: Calculate density

Using the density formula $\rho=\frac{m}{V}$, with $m = 2760\ g$ and $V = 250.0\ cm^{3}$, we have $\rho=\frac{2760\ g}{250.0\ cm^{3}}=11.04\ g/cm^{3}$

Answer:

$11.04\ g/cm^{3}$

Explanation:

Step1: Rearrange density formula for volume

From $\rho=\frac{m}{V}$, we can get $V=\frac{m}{\rho}$. Given $m = 250.0\ g$ and $\rho=0.00134\ g/mL$. $V=\frac{m}{\rho}=\frac{250.0\ g}{0.00134\ g/mL}\approx186567.16\ mL$

Answer:

$186567.16\ mL$

Explanation:

Step1: Rearrange density formula for volume

Using $V=\frac{m}{\rho}$, with $m = 36.2\ g$ and $\rho = 1.60\ g/mL$. $V=\frac{36.2\ g}{1.60\ g/mL}=22.625\ mL$

Answer:

$22.625\ mL$

Explanation:

Step1: Use density formula to find mass

From $m=\rho V$, with $\rho = 0.789\ g/mL$ and $V = 150.0\ mL$. $m=\rho V=0.789\ g/mL\times150.0\ mL = 118.35\ g$

Answer:

$118.35\ g$

A.

Explanation:

Step1: Rearrange density formula for volume

Using $V=\frac{m}{\rho}$, with $m = 30.0\ g$ and $\rho = 1.184\ g/mL$. $V=\frac{30.0\ g}{1.184\ g/mL}\approx25.34\ mL$

Answer:

$25.34\ mL$

B.

Explanation:

Step1: Rearrange density formula for volume

Using $V=\frac{m}{\rho}$, with $m = 30.0\ g$ and $\rho = 1.834\ g/mL$. $V=\frac{30.0\ g}{1.834\ g/mL}\approx16.35\ mL$

Answer:

$16.35\ mL$

C.

Explanation:

Step1: Rearrange density formula for volume

Using $V=\frac{m}{\rho}$, with $m = 30.0\ g$ and $\rho = 1.251\ g/mL$. $V=\frac{30.0\ g}{1.251\ g/mL}\approx23.98\ mL$

Answer:

$23.98\ mL$