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
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948 or 9.48 x 10^2
There are two sets of rules for significant figures
• One set for addition and subtraction
• Another set for multiplication and division
You used the set for multiplication and division.
This problem involves addition and subtraction, and the rule is
The number of places after the decimal point in the answer must be <em>no greater than the number of decimal places in every term</em> in the sum.
Thus, we have
78.9
+890.43
-21.
= 948.33
The "21" term has the fewest digits after the decimal point (none), so the answer must have no digits after the decimal point.
To the correct answer is 948 = 9.48 x 10^2. It has three significant figures.
Explanation:
Formula to calculate hybridization is as follows.
Hybridization = ![\frac{1}{2}[V+N-C+A]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5BV%2BN-C%2BA%5D)
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
So, hybridization of 
is as follows.
Hybridization = ![\frac{1}{2}[V + N - C + A]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5BV%20%2B%20N%20-%20C%20%2B%20A%5D)
= ![\frac{1}{2}[2 + 2]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B2%20%2B%202%5D)
= 2
Hybridization of is sp. Therefore, is a linear molecule. There will be only two electron groups through which Be is attached.
Similarly, hybridization of 
is calculated as follows.
Hybridization =
= ![\frac{1}{2}[8 + 2]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B8%20%2B%202%5D)
= 5
Therefore, hybridization of is ![sp^{3}d. Therefore, [tex]XeF_{2}](https://tex.z-dn.net/?f=sp%5E%7B3%7Dd.%20Therefore%2C%20%5Btex%5DXeF_%7B2%7D)
is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.
Thus, we can conclude that out of the given options is the correct examples of linear molecules for five electron groups.
Answer: 1.Sulfuric acid is a catalyst
2. Vanadium(v) oxide is a catalyst
Answer:
Explanation:
First, find the mass of empirical formula, CH. 12.01 g/mol is for carbon, and 1.008 g/mol is for hydrogen. 12.01+1.008=13.018 G/mol CH. Divide 78.110 G/mol by 13.018 g/mol. You get approximately 6. Multiply that by the subscript of each element. 6(CH)=
