Answer:
9 × 10⁻³ mol·L⁻¹s⁻¹
Explanation:
Data:
k = 1 × 10⁻³ L·mol⁻¹s⁻¹
[A] = 3 mol·L⁻¹
Calculation:
rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹
If we have 6.68% NaClO, it is the same as saying--> 6.68 grams NaClO= 100 mL of solution. we can use this as a conversion.
800. mL (6.68 mL/ 100 mL)= 53.4 mL
solution = solute + solvent
solute= NaClO
solvent= H2O
solvent= 800-53.4= 747 mL of H2O
so, we you need 53.4 mL of NaClO and 747 mL of water or 53.4 grams of NaClO and 747 mL of water
<u>Answer:</u>
<u>For A:</u> The 
for the given reaction is 
<u>For B:</u> The 
for the given reaction is 1642.
<u>Explanation:</u>
The given chemical reaction follows:
The expression of for the above reaction follows:
We are given:
Putting values in above equation, we get:
Hence, the for the given reaction is
Relation of with is given by the formula:
where,
= equilibrium constant in terms of partial pressure =
= equilibrium constant in terms of concentration = ?
R = Gas constant = 
T = temperature = 500 K
= change in number of moles of gas particles = 
Putting values in above equation, we get:
Hence, the for the given reaction is 1642.
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