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Natalija [7]
4 years ago
12

Density of Unknown Substance

Chemistry
1 answer:
8090 [49]4 years ago
3 0
I need points sorry
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Calculate the number of atoms per unit cell in each type of cubic unit cell. Enter three integers separated by commas (e.g., 1,2
agasfer [191]

Answer:

1, 2, 4

Explanation:

The primitive cubic unit cell has eight corners, and each corner is shared with 8 cubic unit cells. Therefore since we have one atom at each corner, we will have:

8 corner x  1 atom/8 corner = 1 atom

There is then the equivalent of one atom per unit cell in the primitive cubic cell.

For the body-centered cubic unit cell, we have 8 corners again shared each by 8 lattices as in the simple cubic plus we have one atom  in the center of the cubic lattice. Therefore, the number of atoms in the body -centerd unit cell is two:

8 corner x 1 atom/8 corner = 1 atom

                      +

1 atom in the center

= 2 atoms/ unit cel

For the face-centerd cubic again we have 8 atoms in the corners shared by 8 lattices, plus 1 atom in each of the faces shared by two unit cells:

8 corner x 1/8 atom/corner + 6 faces x 1 atom/face = 4 atoms/unit cell

4 0
4 years ago
A benefit of using nuclear energy could be the abundance of uranium in nature. Which human risk about mining is the most importa
Murrr4er [49]
The health hazards because it is asking which human risk is the most important to consider.
5 0
3 years ago
Read 2 more answers
746mmHg, 115m^3, 21*C number of moles?
victus00 [196]

Answer:

n = 4678.13 mol

Explanation:

Given data:

Temperature = 21°C

Pressure = 746 mmHg

Volume = 115 m³

Number of moles = ?

Solution:

21+273 = 294 k

746 /760 = 0.982 atm

115×1000 = 115000 L

PV = nRT

n = PV/RT

n =   0.982 atm × 115000 L /0.0821 atm. L. K⁻¹. mol ⁻¹× 294 k

n = 112930 atm. L/24.14 atm. L. mol ⁻¹

n = 4678.13 mol

8 0
3 years ago
What is the correct formula for sulfuric acid
sattari [20]
The correct formula is H20S4
3 0
3 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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