Answer:
Option "B" is correct.
Explanation:
According to VSEPR theory, There are repulsion forces exists among the bond pair - bond pair or bond pair - lone pair of electrons. In 
and 
, the number of electron pairs are same but methane has all the four bond pairs where in ammonia, three bond pairs and one lone pair exists. And thus there are repulsion forces possible in between the lone pair and bond pair of electrons thus the arrangement of electron pairs around both the molecules is same or different depending up on the conditions leading to maximum repulsion.
Explanation:
The observation of student was that thermometer reading changed from 27°C to 35°C which indicates that temperature of the beaker solution rose after reaction due to release of heat during reaction as a product.
Those chemical reactions which gives heat energy as a product into their surrounding are categorized as exothermic reactions. During the course of these reaction temperature of the surroundings also increased.
So, this means that reaction between silver nitrate and copper wire is an exothermic reaction.
Three ways that the student could speed up the reaction :
- By adding catalyst to the reaction.
- By decreasing the temperature.
- By increasing the concentration of silver nitarte solution.
Given:
7.20 g sample of Al2(SO4)3
Required:
Mass of oxygen
Solution:
Since you are not given a
chemical reaction, just base your solution to the chemical formula given.
Molar mass of Al2(SO4)3 = 342.15 g/mol
7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol
O2)(32g O2/1mol O2) = 4.04 g O2
Answer:
a. both temperature changes will be the same
Explanation:
When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:
Q = m×C×ΔT
<em>Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.</em>
Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.
m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.
And Q is the heat released: If 2g release X heat, 4g release 2X.
Thus, ΔT in the experiments is:
Experiment 1:
X / 102C = ΔT
Experiment 2:
2X / 204C = ΔT
X / 102C = ΔT
That means,
<h3>a. both temperature changes will be the same</h3>
Hey there!:
From the given data ;
Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )
Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:
[E]t in molar = g/L * mol/g
[E]t = 0.01 g/L * 1 / 45,000
[E]t = 2.22*10⁻⁷
Vmax = 0.758 umole/min/ per mL
= 758 mmole/L/min
=758000 mole/L/min => 758000 M
Therefore :
Kcat = Vmax/ [E]t
Kcat = 758000 / 2.2*10⁻⁷ M
Kcat = 3.41441 *10¹² / min
Kcat = 3.41441*10¹² / 60 per sec
Kcat = 5.7*10¹⁰ s⁻¹
Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹
Hope that helps!
