The time required to reduce the concentration from 0.00757 M to 0.00180 M is equal to 1.52 × 10⁻⁴ s. The half-life period of the reaction is 9.98× 10⁻⁵s.
<h3>What is the rate of reaction?</h3>
The rate of reaction is described as the speed at which reactants are converted into products. A catalyst increases the rate of the reaction without going under any change in the chemical reaction.
Given the initial concentration of the reactant, C₀= 0.00757 M
The concentration of reactant after time t is C₁= 0.00180 M
The rate constant of the reaction, k = 37.9 M⁻¹s⁻¹
For the first-order reaction: 
0.00180 = 0.00757 - (37.9) t
t = 1.52 × 10⁻⁴ s
The half-life period of the reaction: 

Half-life of the reaction = 9.98 × 10⁻⁵s
Learn more about the rate of reaction, here:
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Answer: yo sorry this a hard one
Explanation:
bro
Answer:
x = 1, -7.5
Explanation:
2x² + 13x = 15
Divide the left side of the equation by 2
2(x² + 6.5x) = 15
Divide 6.5 by 2 and square the quotient
6.5/2 = 3.25
3.25² = 10.5625
Add 10.5625 to the left side
2(x² + 6.5x + 10.5625) = 15
Since you have a 2 outside the parentheses, you will be adding 10.5625•2 to the right side.
10.5625 • 2 = 21.125
2(x² + 6.5x + 10.5625) = 36.125
To factor (x² + 6.5x + 10.5625), add b/2 to x
b/2 = 6.5/2 = 3.25
2(x + 3.25)² = 36.125
Divide by 2
(x + 3.25)² = 18.0625
Square root.
(x + 3.25) = √18.0625
x + 3.25 = ±4.25
Subtract 3.25.
x = 4.25 - 3.25 = 1
x = -4.25 - 3.25 = -7.5
x = 1, -7.5
Answer:
D. Scientific notation is a way of writing large and small numbers
Answer:
The heat absorbed by the sample of water is 3,294.9 J
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
- Q=?
- m= 45 g
- c= 4.184

- ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 C
Replacing:
Q= 4.184
* 45 g* 17.5 C
Solving:
Q=3,294.9 J
<u><em>The heat absorbed by the sample of water is 3,294.9 J</em></u>
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