Answer:
The result is a superposition which is twice the amplitude of each input wave. Φ = π means the two waves are completely OUT OF PHASE, and so add completely destructively. The result is a superposition which has no amplitude at all.
Explanation:
The result is a superposition which is twice the amplitude of each input wave. Φ = π means the two waves are completely OUT OF PHASE, and so add completely destructively. The result is a superposition which has no amplitude at all.
Answer:
20cm^2
Explanation:
Here, Density= Mass/ Volume
=100/5
= 20 cm^2
<span>Determine the root-mean-square sped of CO2 molecules that have an average Kinetic Energy of 4.21x10^-21 J per molecule. Write your answer to 3 sig figs.
</span><span>
E = 1/2 m v^2
If you substitute into this formula, you will get out the root-mean-square speed.
If energy is Joules, the mass should be in kg, and the speed will be in m/s.
1 mol of CO2 is 44.0 g, or 4.40 x 10^1 g or 4.40 x 10^-2 kg.
If you divide this by Avagadro's constant, you will get the average mass of a CO2 molecule.
4.40 x 10^-2 kg / 6.02 x 10^23 = 7.31 x 10^-26 kg
So, if E = 1/2 mv^2
</span>v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 115184.68
Take the square root of that, and you get the answer 339 m/s.
<h3>
Answer:</h3>
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂
[Given] 20 mol Al₂O₃
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol Al₂O₃ → 4 mol Al
<u>Step 3: Stoich</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4:Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
Since our final answer already has 1 sig fig, there is no need to round.
Answer:
Q.1
Given-
Volume of solution-1 L
Molarity of solution -6M
to find gms of AgNO3-?
Molarity = number of moles of solute/volume of solution in litre
number of moles of solute = 6×1= 6moles
one moles of AgNO3 weighs 169.87 g
so mass of 6 moles of AgNO3 = 169.87×6=1019.22
so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution
