Ask question

Ask question

All categories

3 years ago

13

You are given a cube with the length of 2.5 cm , a width of 2.5 cm , and a height of 2.5 cm. You place it on the scale, and its

mass is 295 g. Calculate the density. Please show your work.

1 answer:

kakasveta [241]3 years ago

8 0

Density is equal to the mass divided by the voulme.
So: 2.5 multiplied by 2.5 by 2.5.
Then: 295 divided by 16.625=18.88g/cm to the power 3

You might be interested in

Write a balanced chemical equation for the standard formation reaction of gaseous formaldehyde CH2O

Free_Kalibri [48]

From the combustion of octane, the formaldehyde will be formed as this equation:

C8H18 + O2 → CH2O + H2O this is the original equation but it is not a balanced equation, so let's start to balance it:

the equation to be balanced so the number of atoms on the right side of the equation sholud be equal with the number of atoms on the lef side.

-we have 8 C atoms on left side and 1 atom on the right side so we will try putting 8 CH2O on the right side instead of CH2O

C8H18 + O2 → 8 CH2O + H2O

we have 2 O atoms on the left side and 9 atoms on the right side so we will try first to put 9 O2 instead of O2 on the left side and put 2H2O on the right side and put 16 CH2O instead of 8 CH2O to make the atoms of O are equal on both sides = 18 atoms

C8H18 + 9 O2 → 16 CH2O + 2H2O

put now we have 8 atom C on the left side and 16 atom on the right side so, we will put 2 C8H18 instead of C8H18 now we get this equation:

2C8H18 + 9O2 →16 CH2O + 2H2O

-now we have 36 of H atoms on both sides.

- and 16 of C atoms on both sides.

- and 18 of O atoms on both sides.

now all the number of atoms of O & C & H are equal on both sides

∴ 2C8H18 + 9O2 → 16 CH2O + 2 H2O

is the final balanced equation for the formation of formaldehayde

5 0

3 years ago

Gallium is a metallic element in Group III. It has similar properties to aluminium.

Liula [17]

Answer:

Gallium has Orthorhombic and forms ionic bonds with metals

Explanation:

The structure of gallium is Orthorhombic and it has a high neutron capture cross section. It is very stable in dry air but oxidizes in moist air. It reacts with oxygen to form Ga2O3. It has a valence of III. Its electronic configuration is

Ar 3d10 4s2 4p1

It is found in the 3A group of periodic table and it mostly forms ionic bonds with metal.

3 0

3 years ago

Is bubbles forming in water a chemical change

Stolb23 [73]

It is a physical change

7 0

3 years ago

Read 2 more answers

Explain that : how we explore different between Licl and Nacl

Leona [35]

Answer:

See the explanation below, please.

Explanation:

In the bunsen burner, the gas and air inlet can be regulated manually. In the case of metals (such as lithium and sodium in this case) they contain an electron in the latter in its external electronic configuration. They are characterized by transferring electrons easily and produce the emission of light when excited, producing flames of different colors in the lighter (orange for sodium and red / scarlet for lithium)

8 0

3 years ago

5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time

11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}$

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142$

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen = $d_{O_2}=\frac{1 m}{t}$

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane= $d_{CH_4}=\frac{y }{t}$

$\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142$

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0

3 years ago