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3 years ago

Each degree on the Kelvin scale equals:

Chemistry

2 answers:

borishaifa [10]3 years ago

8 0

A change of 1 Kelvin is exactly the same as a change of 1 degree Celsius.

rjkz [21]3 years ago

3 0

Answer is: 1°C.

A change of 1 Kelvin is the same as a change of 1 degree Celsius.

The temperature T in degrees Celsius (°C) is equal to the temperature T in Kelvin (K) minus 273,15: T(°C) = T(K) - 273.15.

For example:

T(He) = 4,2 K.

T(He) = 4,2 K - 273,15.

T(He) = -268,95°C.

The Celsius scale was based on 0°C for the freezing point of water and 100°C for the boiling point of water at 1 atm pressure.

You might be interested in

Calculate how many grams of o2(g) can be produced from heating 87.4 grams of kclo3(s).

Rufina [12.5K]

<span>34.2 grams Lookup the atomic weights of the involved elements Atomic weight potassium = 39.0983 Atomic weight Chlorine = 35.453 Atomic weight Oxygen = 15.999 Molar mass KClO3 = 39.0983 + 35.453 + 3 * 15.999 = 122.5483 g/mol Moles KClO3 = 87.4 g / 122.5483 g/mol = 0.713188188 mol The balanced equation for heating KClO3 is 2 KClO3 = 2 KCl + 3 O2 So 2 moles of KClO3 will break down into 3 moles of oxygen molecules. 0.713188188 mol / 2 * 3 = 1.069782282 mols So we're going to get 1.069782282 moles of oxygen molecules. Since each molecule has 2 atoms, the mass will be 1.069782282 * 2 * 15.999 = 34.23089345 grams Rounding the results to 3 significant figures gives 34.2 grams</span>

4 0

3 years ago

21. Solve the simultaneous equations graphically taking the values of

tigry1 [53]

Answer:

take the l my gang tfdfhngtyhggggggfggg

3 0

3 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago

Draw the best Lewis structure for NH3 by filling in the bonds, lone pairs, and formal charges. (Assign bonds, lone pairs, radica

kiruha [24]

Answer : The Lewis-dot structure of $NH_3$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $NH_3$

As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in $NH_3$ = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on N}=5-2-\frac{6}{2}=0$

$\text{Formal charge on }H_1=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_2=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_3=1-0-\frac{2}{2}=0$

Hence, the Lewis-dot structure of $NH_3$ is shown below.

3 0

2 years ago

What observation would you look for to tell if a reaction was endothermic?

horsena [70]

Exothermic reactions release heat into their surroundings and endothermic reactions absorb heat and are cool. I believe the correct answer is D.

5 0

1 year ago