1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
almond37 [142]
3 years ago
7

A materials ability to allow heat to flow is called?

Chemistry
2 answers:
777dan777 [17]3 years ago
8 0
The answer is malleability

leva [86]3 years ago
4 0

Conductivity is the correct answer... Not "malleability"

You might be interested in
Convert 22.4 kg/L to kg/mL
liberstina [14]
1 kg/L -------------- 0.001 kg/mL
22.4 kg/L --------- ??

22.4 x 0.001 / 1 => 0.0224 kg/mL
7 0
4 years ago
You are given the drawing of 2 waves. Notice, wave A is taller and Wave B is thinner. Since wave A is taller, wave
nekit [7.7K]
Wave "B" is thinner. Is this the whole question?
4 0
3 years ago
Read 2 more answers
A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L.A. If the contai
marta [7]

Answer:

Approximately 6.81 × 10⁵ Pa.

Assumption: carbon dioxide behaves like an ideal gas.

Explanation:

Look up the relative atomic mass of carbon and oxygen on a modern periodic table:

  • C: 12.011;
  • O: 15.999.

Calculate the molar mass of carbon dioxide \rm CO_2:

M\!\left(\mathrm{CO_2}\right) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

Find the number of moles of molecules in that 41.1\;\rm g sample of \rm CO_2:

n = \dfrac{m}{M} = \dfrac{41.1}{44.009} \approx 0.933900\; \rm mol.

If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:

P \cdot V = n \cdot R \cdot T,

where

  • P is the pressure inside the container.
  • V is the volume of the container.
  • n is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
  • R is the ideal gas constant.
  • T is the absolute temperature of the gas.

Rearrange the equation to find an expression for P, the pressure inside the container.

\displaystyle P = \frac{n \cdot R \cdot T}{V}.

Look up the ideal gas constant in the appropriate units.

R = 8.314 \times 10^3\; \rm L \cdot Pa \cdot K^{-1} \cdot mol^{-1}.

Evaluate the expression for P:

\begin{aligned} P &=\rm \frac{0.933900\; mol \times 8.314 \times 10^3 \; L \cdot Pa \cdot K^{-1} \cdot mol^{-1} \times 298\; K}{3.4\; L} \cr &\approx \rm 6.81\times 10^5\; Pa \end{aligned}.

Apply dimensional analysis to verify the unit of pressure.

4 0
4 years ago
2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an
Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression d=\frac{m}{V} that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL

d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL

d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL

d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL

The problem says that all the samples have the same mass, so:

m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

V_{lithium}=\frac{m}{d_{lithium}}

V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m

V_{lithium}=1.88\frac{mL}{g}*m

V_{gold}=\frac{m}{d_{gold}}

V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m

V_{gold}=5.18*10^{-2}\frac{mL}{g}*m

V_{aluminum}=\frac{m}{d_{aluminum}}

V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m

V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m

V_{lead}=\frac{m}{d_{lead}}

V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m

V_{lead}=8.85*10^{-2}\frac{mL}{g}*m

If we assume m = 1g, we find that:

V_{lithium}=1.88mL

V_{gold}=5.18*10^{-2}mL

V_{aluminum}=3.70*10^{-1}mL

V_{lead}=8.85*10^{-2}mL

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0
3 years ago
n a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the ________, while the antibonding
Kruka [31]

Answer:

In a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the valence band, while the antibonding orbitals that are completely empty are referred to as the conduction band.

The conduction band occupies a higher energy level than the valence band. The band gap is what separates the two orbitals.

3 0
2 years ago
Other questions:
  • How many particles would be found in a 1.224 g sample of K2O
    11·2 answers
  • How many lone pairs of electrons in lewis structure of CH3 OH
    6·1 answer
  • I WILL GIVE THE BRAINIEST ANSWER TO THE FIRST PERSON TO ANSWER.TIMED QUIZ!!!ANSWER QUICKLY PLEASE.
    12·2 answers
  • A 58.0mL sample of a 0.122M potassium sulfate solution is mixed with 40.0mL of a 0.102M Lead(II) acetate solution and this preci
    11·2 answers
  • Which one of the following electron transitions does not correspond to an absorbtion of a quanta of light?
    6·1 answer
  • In what way would one dozen elephants and one dozen doughnuts be different?
    8·1 answer
  • Which series of energy transformations occurs as an electric generator powers a light bulb
    9·2 answers
  • All electromagnetic waves travel at the same speed in a vacuum, usually referred to as the speed of light. This speed is approxi
    15·1 answer
  • What could we use to measure the energy of particles when we are unable to see them
    15·1 answer
  • Listed following are some distinguishing characteristics of comets, meteors, and asteroids. Match these to the appropriate categ
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!