What is the mass of 9.45 mol of aluminum oxide?

A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is 10.59 nm . What is the total w

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Whenever objects are placed in a line with spaced in between them, the number of spaces is one less than the number of objects. For reference, you may hold up four fingers and count the spaced between your fingers.

Therefore, if there are 105 molecules, then there are 104 spaces present.

We also know that 1 mm = 1000 nm. So, we first convert the spacing to millimeters as the final answer is required in millimeters

spacing = 10.59/1000 = 0.01059 mm

Now, the width of the lattice will be (ignoring the size of the molecules):

104 * 0.01059

1.10 millimeters

8 0

4 years ago

Which of the following can be achieved by using a visible-light spectrophotometer?

gulaghasi [49]

1 and 2 i think this is the answer
thnx u

5 0

3 years ago

Determine the mass of carbon iv oxide ,produced on burning 104g of ethyne

mars1129 [50]

163 grams (3 sig. fig.).

<h3>Explanation</h3>

Formula of carbon(IV) oxide (a.k.a. carbon dioxide): $\text{CO}_2$ .
Molar mass of $\text{CO}_2$ : $\underbrace{12.01}_{\text{C}} + 2\times\underbrace{16.00}_{\text{O}}=44.01\;\text{g}\cdot\text{mol}^{-1}$ .

Formula of ethyne: structural $\text{H}-\text{C}\equiv\text{C}-\text{H}$ or molecular $\text{C}_2\text{H}_2$ .
Molar mass of $\text{C}_2\text{H}_2$ : $2\times\underbrace{12.01}_{\text{C}}+2 \times\underbrace{16.00}_{\text{O}} = 56.02\;\text{g}\cdot\text{mol}^{-1}$ .

All carbon atoms in that 104 grams of ethyne will end up in $\text{CO}_2$ . Number of moles of molecules in 104 grams of ethyne:

$n = \dfrac{m}{M} = \dfrac{104}{56.02} = 1.85648\;\text{mol}$ .

There are two carbon atoms in each ethyne molecule. Number of carbon atoms in that many ethyne molecules:

$n(\text{C}) = 2\;n(\text{C}_2\text{H}_2) = 3.71296\;\text{mol}$ .

There are one carbon atom in each $\text{CO}_2$ molecule. In case oxygen is in excess, all those carbon atoms from that 104 grams of ethyne will make $n(\text{CO}_2) = n(\text{C}) =3.71296\;\text{mol}$ of $\text{CO}_2$ .

Mass of all those $\text{CO}_2$ molecules:

$m = n\cdot M = 163\;\text{g}$ . (3 sig. fig. as in the mass of ethyne.)

7 0

4 years ago

Which of the following would be part of an aquatic ecosystem?

Masja [62]

Answer:

Aquatic ecosystem has two components -

Biotic components

2.Abiotic components

temperature and amount of sunlight are the part of abiotic component .

while living things like sponges and planktons are the biotic components of ecosystem.

Explanation:

aquatic components are of two types-

freshwater ecosystem( lakes and ponds, river and streams)

marine ecosystem(ocean ecosysyem, estuaries)

planktons-

planktons are found in limnetic zone, availability of sunlight is much here. planktons are zooplanktons and phytoplanktons are very important link in aquatic ecosystem.

sponges

In marine water, the benthic zone is the area below the pelagic zone. Here temperature decreased because of less light perception. This zone is very nutrient rich so organisms which are present here are- bacteria, fungi, sea anemone, sponges and some fishes.

3 0

3 years ago

Read 2 more answers

If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

Consider the following reaction: 

SO2Cl2 -----> SO2(g) + Cl2(g) 

Kc= 2.99 x 10^-7 at 227 degrees celcius 

If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

SOCl2 ---------> SO2 + Cl2 Kc = 2.99x10⁻⁷

i) 0.168 0 0

c) -x +x +x

e) 0.168-x x x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

[Cl₂] = 2.24x10⁻⁴ M

5 0

3 years ago