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aksik [14]
3 years ago
7

A compound made of elements A and B, has a cubic unit cell. There is an A atom at each corner of the cube and an A atom at the c

enter of each face of the cube. There are four B atoms that lie entirely within the unit cell. Based on this information, the empirical formula for the compound is:
Chemistry
1 answer:
Lorico [155]3 years ago
5 0

Answer:

A₅B₄

Explanation:

Since we have one atom of element A at the center of each face of the unit cell, since the unit cell is a cubic cell, we have 6 faces. Since the atom on the face of the unit cell is shared with another cell, we have half of it in the unit cell is shared So, the number of atoms per face is 1/2 atom/face × 6 faces = 4 atoms on the faces of the unit cell.

Also, we have 1 atom at each corner of the cubic unit cell. Since there are 8 corner in the cubic unit cell. Also, each atom at the corner is shared with 8 unit cells, so we have 1/8 atom per corner. So, the number of atoms per unit cell is 1/8 atom/corner × 8 corners = 1 atoms at the corners of the unit cell.

So, in total we have 4 + 1 = 5 atoms of element A in the unit cell.

Also, there are 4 atoms of element B in the unit cell.

So, the ratio of atoms of element A to element B is 5 : 4.

A:B = 5:4

So, the empirical formula of the compound containing elements A and B is A₅B₄

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In a constant-pressure calorimeter, 55.0 mL of 0.340 M Ba(OH), was added to 55.0 mL of 0.680 M HCI. The reaction caused the temp
Brut [27]

Answer:

Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O

The reactants are present in equimolar amounts, so there is no excess or limiting reactants.

(0.0500 L) x (0.600 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0300 mol H2O

(4.184 J/g·°C) x (50.0 g + 50.0 g) x (25.82 - 21.73)°C = 1711.256 J

(1711.256 J) / (0.0300 mol H2O) = 57042 J/mol = 57.0 kJ/mol H2O

Explanation:

4 0
3 years ago
If a sample contains 70.0 % of the R enantiomer and 30.0 % of the S enantiomer, what is the enantiomeric excess of the mixture?
NeTakaya

Answer:

the enantiomeric excess of the mixture is 40%

Explanation:

The computation of the enantiomeric excess of the mixture is shown below:

As we know that

= |\frac{R - S}{R + S} |\times 100\\\\= |\frac{70 - 30}{70 + 30}| \times 100\\\\= 40\%

Hence, the enantiomeric excess of the mixture is 40%

3 0
3 years ago
Scientists excavating at the bottom of the ocean come across fossilized remains of an ancestor of a mountain goat. How can this
galina1969 [7]
There was land on that surface before there was water
4 0
3 years ago
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the max
jolli1 [7]

<u>Answer:</u> The mass of aluminium chloride that can be formed are 46.3 g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  ....(1)  

  • <u>For Aluminium:</u>

Given mass of aluminium = 32 g  

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:  

\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol

  • <u>For Chlorine:</u>

Given mass of chlorine = 37 g  

Molar mass of chlorine = 71 g/mol

Putting values in above equation, we get:  

\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol

For the given chemical equation:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

By Stoichiometry of the reaction:

3 moles of chlorine gas is reacting with 2 moles of aluminium.

So, 0.521 moles of chlorine gas will react with = \frac{2}{3}\times 0.521=0.347moles of aluminium.

As, given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

So, chlorine gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of chlorine gas is producing 2 moles of aluminium chloride

So,  0.521 moles of chlorine gas will react with = \frac{2}{3}\times 0.521=0.347moles of aluminium chloride.

Now, calculating the mass of aluminium chloride by using equation 1, we get:

Moles of aluminium chloride = 0.347 moles

Molar mass of aluminium chloride = 133.34 g/mol

Putting all the values in equation 1, we get:

0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g

Hence, the mass of aluminium chloride that can be formed are 46.3 g

7 0
4 years ago
At what temperature does a solid turn into a liquid
Lostsunrise [7]

Answer:

0°C

<h3>Hope this helps </h3><h3>Do mark as brainliest </h3>

6 0
4 years ago
Read 2 more answers
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