Answer:
0.025 moles of NO₂ will produced
Explanation:
Given data:
Moles of NO₂ formed = ?
Volume of HNO₃ = 25.0 mL
Molarity of HNO₃ = 2 M
Solution:
Chemical equation:
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Number of moles of HNO₃:
Molarity = number of moles / volume in L
2M = number of moles / 0.025 L
Number of moles = 2 M × 0.025 L
Number of moles = 0.05 mol
Now we will compare the moles of HNO₃ with NO₂ from balance chemical equation.
HNO₃ : NO₂
4 : 2
0.05 : 2/4×0.05 =0.025
0.025 moles of NO₂ will produced.
Explanation:
Mitosis and Meiosis are forms of cell division. They use the same steps for cell division which are Prophase, Metaphase, Anaphase and Telophase.
Mitosis produces 2 diploid cells while Meiosis produces 4 haploid cells.
Meiosis stages -
- Interphase
- Prophase I
- Metaphase I
- Anaphase I
- Telophase I
- Prophase II
- Metaphase II
- Anaphase II
- Telophase II
- Cytokineses
Mitosis stages
- Prophase
- Prometaphase
- Metaphase
- Telophase
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Answer:
fraction of vacancies for this metal FV = 1.918*10⁻⁴
Explanation:
Given:
The number of vacancies per unit volume => ( Nv = 1*10²⁵ m⁻³ )
But we know that Avogrado's constant NA = 6.022*10²³ atoms/mol
Density of the material is given in g/cm3 we need to convert it to g/m³
Density of material ( p ) in g/m³ :
To convert we know that
1 g/cm³ = 1000000 g/m³ then
7.40 * ( 1000000 ) = 7.40*10⁶ g/m³
So, Density of material ( p ) in g/m³ = 7.40*10⁶ g/m³
Given Atomic mass = 85.5 g/mol
To Calculate the number of atomic sites per unit volume , we will use the below formula by substituting those values above
N = NA * p / A
N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5
N = 4.45*10³⁰ / 85.5
N = 5.212*10²⁸ atoms/m³
We can now Calculate the fraction of vacancies using the formula below;
Fv = Nv / N
Fv = 1*10²⁵ / 5.212*10²⁸
fraction of vacancies for this metal at 600c.= 1.918*10⁻⁴
B. The cathode is the electrode where the reduction takes place took test
tides are caused by the gravitational force of the sun and the moon
