Answer:
1.99 x 10⁻¹⁸J
Explanation:
Given parameters:
Frequency of the wave = 3 x 10¹⁵Hz
Unknown:
Energy of the photon = ?
Solution:
To solve this problem, we use the expression below;
E = hf
Where E is the energy, h is the Planck's constant and f is the frequency
Now insert the parameters and solve for E;
E = 6.63 x 10⁻³⁴ x 3 x 10¹⁵ = 19.9 x 10⁻¹⁹J or 1.99 x 10⁻¹⁸J
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
To learn more about stoichiometric quantities visit:
<h3>
brainly.com/question/28174111</h3>
#SPJ4
<h3>
Answer:</h3>
P₂ = 0.67 atm
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Gas Laws</u>
Boyle's Law: P₁V₁ = P₂V₂
- P₁ is pressure 1
- V₁ is volume 1
- P₂ is pressure 2
- V₂ is volume 2
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] P₁ = 2.02 atm
[Given] V₁ = 4.0 L
[Given] V₂ = 12.0 L
[Solve] P₂
<u>Step 2: Solve</u>
- Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
- [Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
- [Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
- [Pressure] Rewrite: P₂ = 0.673333 atm
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>
0.673333 atm ≈ 0.67 atm
D) basic ions contain some H+ ions
1. exercise because you are using exercise to affect the amount of concentration. 2. concentration because concentration is what is being measured