A150 no container can hold 23.6 g of aluminum what is the density of the aluminum?
the density of mercury is 13.6g/cm3 at 23c what is the volume of a container that can hold 3.2 g of mercury at this temperature?
what is the mass of iron that has a density of 4.9g/cm3 in a 250ml beaker?
could you show your work and explain how to do it i'm not sure how to do this

# The density of the aluminum is 59/375 ≅ 0.157 g/cm³

# The volume of the container is 4/17 ≅ 0.235 cm³

# The mass of iron is 1225 g

Step-by-step explanation:

* Lets explain how to solve the problem

- Density is a measurement that compares the amount of matter

 of an object to its volume [ Density = mass/volume]

- Density is found by dividing the mass of an object by its volume

- The unit of measuring density is gram per centimeter³ (milliliter)

* Lets solve the problems

# A container of volume 150 cm³ can hold 23.6 g of aluminum

∵ Density = mass/volume

∵ The mass = 23.6 g

∵ The volume = 150 cm³

∴ The density = 23.6/150 = 59/375 ≅ 0.157 g/cm³

* The density of the aluminum is 59/375 ≅ 0.157 g/cm³

# The density of mercury is 13.6 g/cm³ at 23 C°

- A container can hold 3.2 g of mercury at this temperature

∵ Density = mass/volume

∵ The mass = 3.2 g

∵ The density = 13.6 g/cm³

∴ 13.6 = 3.2/volume

- By using cross multiplication

∴ The volume = 3.2/13.6 = 4/17 cm³

* The volume of the container is 4/17 ≅ 0.235 cm³

# The density of the iron is 4.9 g/cm³

- The beaker has a volume 250 ml

∵ Density = mass/volume

∵ The ml = cm³

∴ The volume of the beaker is 250 cm³

∵ The density = 4.9 g/cm³

∴ 4.9 = mass/250

- Multiply both sides by 250

∴ mass = 1225 g

* The mass of iron is 1225 g

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;