Mercury may have ice trapped in the crater floors near the planet’s north pole.

What is the density of an object that has a mass of 21 g and a volume of 9 ml?

mylen [45]

The formula is density = mass / Volume

density = ?

mass = 21 g

Volume = 9 mL

d = m / V

d = 21 g / 9 mL

d = 2.3 g/mL

Hope this help!

5 0

3 years ago

Isotopes with unstable nuclei are_______and are called _______.

Inga [223]

Answer:

Radioactive and are often called radioisotopes

8 0

3 years ago

How can a substance stay at a certain temperature even though it is being heated?

SOVA2 [1]

The temperature of a certain substance can be seen as the average speed of the atoms or molecules in that substance. In the liquid state of a substance the forces between the atoms or molecules are strong enough to keep them together, however with enough freedom to move, unlike in the solid state. If we would have a closer look at the surface of a liquid from sideways, we would see water molecules jumping out of the water and reentering it again. The lower the water temperature would be the lesser the amount of water molecules leaving the liquid phase would be. If water would be heated up and the temperature will reach 100 degrees C at normal atmospheric pressure, more water molecules would leave the water than reentering. Boiling has started. The temperature of the water remains at 100 degrees C, if the heating continues as the average speed of molecules will not increase, only the rate of molecules leaving the water will increase, until all the water in liquid state has been vapourized. The amount of heat needed to vapourize liquid water is called latent heat. Latent heat is a very important driving factor in the atmosphere and thus the weather.

5 0

3 years ago

In a solution of a carbonated beverage, what is the dissolved carbon dioxide?

insens350 [35]

Answer:

D.

Explanation:

D is the correct answer because, in aqueous solution, solvent is water and solute (in this example carbon dioxide CO₂) is a substance dissolved in water. The amount of solute that can be dissolved in a solvent depends of chemical composition, temperature and pressure

4 0

3 years ago

Read 2 more answers

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

4 years ago