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e-lub [12.9K]
3 years ago
13

Silicon crystals are semiconductors. Which of the following is a correct reason for the increase in the conductivity of Si cryst

als when a small fraction of Si atoms are replaced with those of a different element?
A: P atoms introduce additional mobile negative charges.
B: P atoms introduce additional mobile positive charges.
C: Ge atoms have more electrons than Si atoms have.
D: Ge atoms are much smaller than Si atoms.
Chemistry
2 answers:
Flura [38]3 years ago
8 0

Answer:

c

Explanation:

it's correct I think and hope

melamori03 [73]3 years ago
3 0

Answer:

A is the right answer.. .............

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Which question cannot be answered using scientific methods?
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The question that cannot be answered using scientific method is "Which is the most interesting acid?" 
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1. Define the units for molarity. Page |152 Exp. 14 pH and Solutions Pre Lab Questions (1 of 1) 2. A solution contains 5.25 gram
Ne4ueva [31]

Answer:

1. mol/L

2. 0.120 M

Explanation:

1. Molarity is equal to the moles of solute divided by the liters of solution. The units of molarity are mol/L.

2.

Step 1: Given data

  • Mass of sodium chloride (solute): 5.25 g
  • Volume of solution (V): 750.0 mL = 0.7500 L

Step 2: Calculate the moles of solute (n)

The molar mass of NaCl is 58.44 g/mol.

n = 5.25 g × 1 mol/58.44 g = 0.0898 mol

Step 3: Determine the molarity of the solution

We will use the definition of molarity

M = n/V

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7 0
3 years ago
A 35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base
Gekata [30.6K]
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
                           = 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
                           CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial                  0.00525                0                         0
change             - 0.002625         +0.002625     +0.002625
equilibrium      0.002625             0.002625        0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
                 = -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
      = 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
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3 years ago
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