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amid [387]
3 years ago
10

Why is copper a good electrcity conducter

Chemistry
2 answers:
sineoko [7]3 years ago
6 0

Answer:

Each copper atom has lost one electron and become a positive ion. ... The electrons can move freely through the metal. For this reason, they are known as free electrons. They are also known as conduction electrons, because they help copper to be a good conductor of heat and electricity.

Anvisha [2.4K]3 years ago
3 0

Explanation:

<h3>HOPE IT WILL HELP YOU TO SOLVE YOUR PROBLEM.</h3>

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0.446 g of hydrogen gas fills a 5.0 L bag determine the density of hydrogen
Oduvanchick [21]

The density of hydrogen : ρ = 0.0892 g/L

<h3>Further explanation</h3>

Given

mass of Hydrogen : 0.446 g

Volume = 5 L

Required

The density

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

The unit of density can be expressed in g/cm³, kg/m³, or g/L

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

Input the value :

ρ = m : V

ρ = 0.446 g : 5 L

ρ = 0.0892 g/L

5 0
3 years ago
What are two ways in which white blood cells fight pathgons that have entered the body​
crimeas [40]

Answer:

Answer c i believe

Explanation:

3 0
3 years ago
Calculate the speed of a dog running through a field if he is covering 23.7 meters in 54 seconds.
Elenna [48]

Explanation:

speed = distance/time

= 23.7/54 m/s

= 0.44 m/s

speed of a dog running through a field = 0.44 m/s

4 0
3 years ago
Read 2 more answers
A 50.0 g sample of scandium, sc, is heated by exposure to 1.50 x 10 3 j. The temperature of the sc is raised by 61.1 o
statuscvo [17]

Given mass of Scandium = 50.0 g

Increase in temperature of the metal when heated = 61.1^{0}C

Heat absorbed by Scandium = 1.50*10^{3}J

The equation showing the relationship between heat, mass, specific heat and temperature change:

Q = m C (deltaT)

Where Q is heat = 1.50*10^{3}J

m is mass = 50.0 g

ΔT = 61.1^{0}C

On plugging in the values and solving for C(specific heat) we get,

1.50*10^{3}J=50.0g(C)(61.1^{0}C)

C = 0.491\frac{J}{g^{0}C }

Specific heat of the metal = 0.491\frac{J}{g^{0}C }

7 0
3 years ago
Biacetyl, the flavoring that makes margarine taste "just like butter," is extremely stable at room temperature, but at 200°C it
Sholpan [36]

Answer:

3.91 minutes

Explanation:

Given that:

Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;

As we known that the half-life for first order is:

t__{1/2}}= \frac{0.693}{k}

where;

k = constant

The formula can be re-written as:

k = \frac{0.693}{t__{1/2}}

k = \frac{0.693}{9.0 min}

k = 0.077 min^{-1}

Let the initial amount of butter flavor in the food be (N_0) = 100%

Also, the amount of butter flavor retained at 200°C (N_t)= 74%

The rate constant k = 0.077 min^{-1}

To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:

\frac{N_t}{N_0}= -kt

t = - (\frac{1}{k}*In\frac{N_t}{N_0}  )

Substituting our values; we have:

t = - (\frac{1}{0.077}*In\frac{74}{100}  )

t = 3.91 minutes

∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes

4 0
3 years ago
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