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3 years ago

What is matellic bonding

1 answer:

soldier1979 [14.2K]3 years ago

4 0

Metallic bonding is the force of attraction between valence electrons and the metal atoms. It is the sharing of many detached electrons between many positive ions, where the electrons act as a "glue" giving the substance a definite structure. It is unlike covalent or ionic bonding.

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By paying low wages, factory owners were able to

sweet [91]

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4 years ago

When two aqueous solutions are mixed together the temperature of the solution decreases?

Andre45 [30]

When two aqueous solutions are mixed together and the temperature of the solution decreases, the reaction is called an endothermic reaction with a positive enthalpy change. Endothermic reaction is when heat is absorbed by the reaction taking place. All endothermic reaction always have a positive enthalpy change.

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3 years ago

Living things are classified into groups based on similar characteristics which of the following best represents how plants are

Inessa05 [86]

Pretty sure the answer is D.

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3 years ago

Carbon monoxide (CO) is a poisonous gas because it binds very strongly to the oxygen carrier hemoglobin in blood. A concentratio

velikii [3]

<u>Answer:</u> The amount of CO that is occupied in the room is $1.98\times 10^3L$

<u>Explanation:</u>

We are given:

Concentration of CO = $8.00\times 10^2ppm=800pm$ by volume

This means that $800\mu L\text{ or }800\times 10^{-6}$ of CO is present in 1 L of blood

To calculate the volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of cuboid

l = length of cuboid = 10.99 m

b = breadth of cuboid = 18.97 m

h = height of cuboid = 11.89 m

$V=10.99\times 18.97\times 11.89=2478.83m^3$

Converting this into liters, by using conversion factor:

$1m^3=1000L$

So, $2478.83m^3=2.479\times 10^6L$

Applying unitary method:

In 1 L of blood, the amount of CO present is $800\times 10^{-6}$

So, in $2.479\times 10^6L$ of blood, the amount of CO present will be = $\frac{800\times 10^{-6}}{1}\times 2.479\times 10^{6}=1983.2L=1.98\times 10^3L$

Hence, the amount of CO that is occupied in the room is $1.98\times 10^3L$

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3 years ago

I need to fix the equations

marishachu [46]

Answer:

try looking the equations or find a calulator for those type of eqautions

Explanation:

3 0

3 years ago