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Semenov [28]
3 years ago
15

What is the IUPAC for the compound ZnO?

Chemistry
1 answer:
brilliants [131]3 years ago
5 0
Zinc Oxide is the IUPAC name for ZnO

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How many moles of gas sample are 5.0 L container at 373K and 203kPa
Rom4ik [11]
For the purpose we will here use t<span>he ideal gas law:

p</span>×V=n×R×<span>T

V= </span><span>5.0 L

T= </span><span>373K

p= </span><span>203kPa
</span><span>
R is </span> universal gas constant, and its value is 8.314 J/mol×<span>K
</span>
Now when we have all necessary date we can calculate the number of moles:

n=p×V/R×T

n= 203 x 5 / 8.314 x 373 = 0.33 mole
 
6 0
3 years ago
A 2.241-g sample of nickel reacts with oxygen to form 2.852 g of the metal oxide.
nlexa [21]

Answer:

The empirical formula is = NiO

Explanation:

Given that:- Mass of nickel = 2.241 g

Mass of the oxide formed = 2.852 g

Mass of the oxygen reacted = Mass of the oxide formed - Mass of nickel  = 2.852 g - 2.241 g = 0.611 g

Molar mass of nickel  = 58.6934 g/mol

Moles of nickel = \frac{2.241}{58.6934}\ mol = 0.03818 mol

Molar mass of oxygen  = 15.999 g/mol

Moles of nickel = \frac{0.611}{15.999}\ mol = 0.03818 mol

Taking the simplest ratio for Ni and O as:

0.03818 : 0.03818 = 1 : 1

<u>The empirical formula is = NiO </u>

4 0
3 years ago
Describe two advantages and two disadvantages of building your 3D model with modeling clay. (5 points)
marishachu [46]

Answer:

A model or simulation is only as good as the rules used to create it. It is very difficult to create an entirely realistic model or simulation because the rules are based on research and past events. The main disadvantage of simulations is that they aren't the real thing.

Explanation:

8 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
Explain what would happen in Step 2 if the cylinder were heated with the plunger held steady.
Svetradugi [14.3K]

Answer:

be more descriptive

Explanation:

4 0
3 years ago
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