All categories

3 years ago

A solid has shape and volume?

Chemistry

1 answer:

tia_tia [17]3 years ago

6 0

Answer:

A solid has a definite volume and shape, a liquid has a definite volume but no definite shape, and there is no definite volume or shape of a gas. Usually, the transition from solid to liquid does not change a substance's volume substantially.

Explanation:

You might be interested in

What is the total number of atoms in 3.0 molecules of propanone, CH3COCH3? (Avogadro’s constant = 6 x 1023 mol-1)

BigorU [14]

Answer:

1.81 x 10²⁴atoms

Explanation:

Given parameters:

Number of molecules = 3 moles

Unknown:

Number of atoms = ?

Solution:

To solve this problem, we are going to use the mole concept.

1 mole of any substance contains 6.02 x 10²³ atoms

3 mole of propanone will contain 3 x 6.02 x 10²³ atoms = 1.81 x 10²⁴atoms

4 0

3 years ago

Which type of solid can be described by the unknown solid if it has the properties listed in the table below?

Bingel [31]

Answer:

Ionic

Explanation:

8 0

3 years ago

Given: glow is a find the indicated measure.

Alisiya [41]

Answer:

is there supposed to be some type of image or ...

Explanation:

4 0

3 years ago

What type of experiment did the student in the passage perform?

Angelina_Jolie [31]

Answer:

controlled

Explanation:

awnser

4 0

3 years ago

Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)

cupoosta [38]

Answer:

$m_{CaCO_3}=0.179gCaCO_3$

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2$

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

$m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3$

Regards.

3 0

4 years ago