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maxonik [38]
2 years ago
5

How many grams of Silicon (Si, 28.09 g/mol) are in a computer chip with 1.08 x 10^23 atoms of Si

Chemistry
2 answers:
mote1985 [20]2 years ago
8 0
1 mol of any particles - 6.02 *10²³

(1.08*10²³ * 1 mol)/(6.02*10²³) = 1.08*10²³/6.02*10²³ mol Si

(1.08*10²³/6.02*10²³) mol Si * 28.09 g/1mol = (28.09*1.08*10²³)/6.02*10²³ ≈
≈ 5.04 g Si

telo118 [61]2 years ago
6 0

Answer: 5.04 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given atoms}}{\text {Avogadro's no}}

For silicon:

Atoms of silicon given = 1.08\times 10^{23}

Avogadro's number = 6.023\times 10^{23}

Putting values in above equation, we get:

\text{Moles of silicon}=\frac{1.08\times 10^{23}}{6.023\times 10^{23}}=0.18mol

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number 6.023\times 10^{23} of particles

1 mole of Si weighs = 28.09 g

0.18 moles of Si contains =\frac{28.09}{1}\times 0.18=5.04g

Thus the mass of silicon in a computer chip is 5.04 grams.

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Carbon monoxide and molecular oxygen react to form carbon dioxide. A 50.0 L reactor at 25.0 oC is charged with 1.00 bar of CO. T
Umnica [9.8K]

Answer:

CO = zero

CO2 =1 bar

O2  = 2.02 bar

Explanation:

We are given

initial pressure of CO = 1bar

total pressure = 3.52 bar

so initial pressure of O2 = 3.52 - 1 = 2.52 bar

the reaction is

2CO + O2 →  2CO2

using the unitary method

2 moles of CO2 → 1 mole of O2

1  bar of CO → \frac{1}{2} * 1= 0.5 bar (required)

but we have more oxygen present , that means CO is the limiting reagent

  • Final pressure of CO will be zero as it is the limiting reagent so it will be consumed completely
  • 1 bar of CO → \frac{2mol CO2}{2mol CO} * 1= 1 bar of CO2
  • 2.52 bar O2 (initially) - 0.5 bar (reacted) = 2.02bar O2
6 0
2 years ago
A human lung at maximum capacity has a volume of 3.0 liters. If the partial pressure of oxygen in the air is 21.1 kilopascals an
MrRa [10]

Answer : 0.026 moles of oxygen are in the lung

Explanation :

We can solve the given question using ideal gas law.

The equation is given below.

PV = nRT

We have been given P = 21.1 kPa

Let us convert pressure from kPa to atm unit.

The conversion factor used here is 1 atm = 101.3 kPa.

21.1 kPa \times \frac{1atm}{101.3kPa}= 0.208 atm

V = 3.0 L

T = 295 K

R = 0.0821 L-atm/mol K

Let us rearrange the equation to solve for n.

n = \frac{PV}{RT}

n = \frac{0.208atm\times 3.0L}{0.0821 L.atm/mol K\times 295 K}

n = 0.026 mol

0.026 moles of oxygen are in the lung

3 0
2 years ago
Which of the following correctly describes a compound? (4 points)
wariber [46]

Explanation:

The atoms are chemically bonded together, and they retain their individual physical and chemical properties.

8 0
2 years ago
A 52.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 26.0 mL of KOH at 25 ∘C.
prohojiy [21]
The balanced equation for the above reaction is 
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL  = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs. 
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
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2 years ago
Plasmid DNA and a gene of interest are cut with the enzyme PpuMI. Write a possible sequence of bases for the sticky end of the g
nasty-shy [4]

Answer:

5' RG        GWCCY 3'

3' YCCWG        GR 5'

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The enzyme PpuMI is a restriction endonuclease enzyme, it has a specific recognition site where it cut the DNA. The source of the enzyme is from ​​an E. coli strain that carries the PpuMI gene from Pseudomonas putida (R. Morgan).

The enzyme PpuMI recognizes specific sequence with palindrome arrangement. It target the sequence 5' RGGWCCY 3'

target Sequence: 5' RGGWCCY 3'

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The enzyme cleavage point is at:

5' RG^GWCCY 3'

3' YCCWG^GR 5'

The product of the cleavage will give a sticky end Cleavage:

5' RG        GWCCY 3'

3' YCCWG        GR 5'

Note: R stands for purines (adenine and guanine). Y stands for pyrimidines (cytosine, thymine, and uracil). And W represents adenine or thymine.

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