What is the density of an unknown object that has a mass of 5.0 g and volume of 20<br> ml?

Leviafan [203]

Common insoluble (sparingly soluble) salts are carbonates, hydroxides<span>, sulfates, and sulfides.</span>

3 0

4 years ago

A container of hydrogen at 172 kPa was decreased to 85.0 kPa producing a new volume of 3L. What was the original volume?

Aneli [31]

Answer:

<h2>The answer is 1.48 L</h2>

Explanation:

In order to find the original volume we use the same for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the original volume

$V_1 = \frac{P_2V_2}{P_1} \\$

From the question

P1 = 172 kPa = 172000 Pa

P2 = 85 kPa = 85000 Pa

V2 = 3 L

We have

$V_1 = \frac{85000 \times 3}{172000} = \frac{255000}{172000} = \frac{255}{172} \\ = 1.482558139...$

We have the final answer as

Hope this helps you

4 0

3 years ago

Water _____.

Sergio039 [100]

Answer: is neither an acid nor a base

Explanation: Water is a universal solvent which means it can dissolve most of the substances in it.

Water has high thermal heat capacity , which means large heat is required to heat the water.

Water is not always pure as it gets contaminated by various pollutants present in the atmosphere such as gases, bacteria and suspended matter.

Water is an amphoteric substance which can act as both acid and base, thus can donate and acept [texH^+[/tex] ions.Thus it is neither an acid nor a base.

$HCl+H_2O\rightarrow H_3O^++Cl^-$

Here water is accepting a proton, thus it acts as base.

$NH_3+H_2O\rightarrow NH_4^++OH^-$

Here water is donating a proton, thus it acts as acid.

4 0

3 years ago

5. An object is a regular, rectangular, solid with dimensions of 2 cm by 3cm by 2cm. It has a mass of 24 g. find its density.

m_a_m_a [10]

Answer:

Density, D = 2g/cm^3

Explanation:

Given the following data;

Length = 2cm

Width = 3cm

Height = 2cm

Mass = 24g

Density = ?

Volume of a rectangular solid (V) = Length × Weight × Height

Therefore, V = L× W × H

Substituting the values, we have;

$V = 2 *3 * 2 = 12$

V = 12cm

Density can be defined as the ratio of mass to volume i.e mass all over volume.

Mathematically, $Density = \frac{Mass}{Volume}$

$D = \frac{M}{V}$

Substituting the values, we have;

$D = \frac{24}{12}$

<em>Density, D = 2g/cm^3</em>

Hence, the density of the rectangular solid is 2g/cm^3.

3 0

3 years ago

What mass (in g) of KIO3 is needed to prepare 50.0 mL of 0.20 M KIO3? b) What volume (in mL) of 0.15 m H2SO4 is needed to prepar

irina1246 [14]

Answer:

a) mass = 2.14 g

b) volume = 14.76 mL

Explanation:

a) mass of KIO3:

In order to know the mass of any compound, we need the molar mass and the moles so:

m = n * MM

to calculate the moles, we already have the concentration of the solution required and it's volume, so, from there, we can calculate the moles:

n = M * V

Replacing the data (And converting mL to L, dividing by 1000):

n = 0.2 * 0.050 = 0.01 moles

Now, the reported molar mass of KIO3 is 214 g/mol so, the mass:

m = 0.01 * 214

<u>m = 2.14 g of KIO3 are needed to prepare this solution</u>

b) volume of H2SO4 needed:

In this case, we have a little issue, the concentration of the H2SO4 given is expressed in molality (m) and not molarity (Capital M), so, in order to convert this to molarity, we need the density. The density of H2SO4 we need to use here is 1.83 g/mL, so, let's convert the molality to molarity and then, the volume:

m = n/kg solvent

Now, 0.15 m we can rewrite this like:

0.15 moles solute / kg solvent

In this case, we have 0.15 moles in 1 kg of solvent or 1000 g (Which we can assume is water).

Now, that we have the moles, let's calculate the mass of acid, with the molecular weight (MM = 98 g/mol)

mass = 0.15 * 98 = 14.7 g of acid.

Now we know that mass of solution = mass solute + mass solvent

mass solution = 14.7 + 1000 = 1014.7 g of solution

With the density, we can calculate the volume of solution:

d = mass/V ---> V = mass / d

V = 1014.7/1.83 = 553.48 mL of solution

So the 0.15 m, are in 553.48 mL of solution, or 0.55348 L of solution, therefore, the molarity will be:

M = 0.15 / 0.55348 = 0.271 M

Now that we have the molarity, for relation of mole ratio, we know that:

moles A = moles B

we will say "A" is the original solution, and "B" the solution needed.

nA = nB

therefore we know that nA is:

nA = Ma*Va

nB = MB * VB

replacing we have:

0.271Va = 0.080 * 50

<u>And this is the volume of acid needed to take to prepare the solution.</u>

8 0

3 years ago

What is the density of an unknown object that has a mass of 5.0 g and volume of 20 ml?