Answer:
1. 23.3 g
2. 82 g
3. 0.402 mol
4. 0.506 mol
Explanation:
We can convert moles (n) to mass (m) using the following expressions that include the molar mass (M).
n = m/M
m = n × M
1. The molar mass of NH₄Cl is 53.49 g/mol.
m = 0.436 mol × 53.49 g/mol = 23.3 g
2. The molar mass of Ca(NO₃)₂ is 164.09 g/mol.
m = 0.50 mol × 164.09 g/mol = 82 g
3. The molar mass of NaCl is 58.44 g/mol.
n = 23.5 g/(58.44 g/mol) = 0.402 mol
4. The molar mass of KMnO₄ is 158.03 g/mol.
n = 79.9 g/(158.03 g/mol) = 0.506 mol
Answer:
Option d is correct option = Frequency of light is 6.56×10⁻² Hz
Explanation:
Given data:
Frequency of light = ?
wavelength of light = 4.57×10⁹m
Solution:
Formula:
Speed of light = wavelength × frequency
c = λ × f
f = c/λ
This formula shows that both are inversely related to each other.
The speed of light is 3×10⁸ m/s
Frequency is taken in Hz.
It is the number of oscillations, wave of light make in one second.
Wavelength is designated as "λ" and it is the measured in meter. It is the distance between the two crust of two trough.
Now we will put the values in formula.
f = 3×10⁸ m/s / 4.57×10⁹m
f = 0.656×10⁻¹s⁻¹
s⁻¹ = Hz
f = 0.656×10⁻¹ Hz or 6.56×10⁻² Hz
Answer:
- 1. Iodine is the limiting reactant
- 2.
Explanation:
<u>1. Balanced chemcial equation:</u>
<u>2. Theoretical mole ratio</u>
It is the ratio of the coefficients of the reactants in the balanced chemical equation:
<u>3. Actual ratio</u>
<u />
It is ratio of the moles available to reat:
<u>4. Comparison</u>
Then, there are more aluminum available than what is needed to react with the 9 moles of iodine, meaning that the aluminum is in excess and the iodine will react completely, being the latter the limiting reactant.
Conclusion: iodine is the limiting reactant.
<u>5. How much aluminum iodide will be produced?</u>
Use the theoretical mole ratio of aluminum iodide to iodide:
Types of Bonds can be predicted by calculating the difference in electronegativity.
If, Electronegativity difference is,
Less than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For C and N,
E.N of Nitrogen = 3.04
E.N of Carbon = 2.55
________
E.N Difference 0.49 (Weakly Polar Covalent)
For N and S,
E.N of Nitrogen = 3.04
E.N of Sulfur = 2.58
________
E.N Difference 0.46 (Weakly Polar Covalent)
For K and F,
E.N of Fluorine = 3.98
E.N of Potassium = 0.28
________
E.N Difference 3.70 (Ionic)
For N and N,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference 0 (Non Polar Covalent)
