Answer:
Zero order
Explanation:
Looking at the data we can note a linear dependence between concentration and time.
Time Conc.
0 2
15 1.82
30 1.64
48 1.42
75 1.10
In the first 15 min it was consumed 2-1.82=0.18. So the rate is 
From 15 to 30 min (it has passed 15 min) is consumed 1.82-1.64=0.18, so as in the previous calculation the rate is 
.
From 30 to 48 (it has passed 18 min)the rate is 
From 48 to 75 (it has passed 27 min) the rate is 
So these results suggest that despite of the ever minor concentration of the reactant the rate is ever the same. Hence the reaction rate could be expressed as 
that is, the reaction is the zero order respect to C2H5I since it is not depending on concentration of C2H5I.
The question is incomplete, the complete question is:
A chemist prepares a solution of vanadium (III) chloride (VCl3) by measuring out 0.40g of VCl3 into a 50.mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl− anions in the chemist's solution. Be sure your answer is rounded to the correct number of significant digits.
Answer:
0.153M of anions
Explanation:
First we calculate the concentration of the solution. From m/M= CV
m=given mass, M= molar mass, C =concentration of solution, V= volume of solution
Molar mass of compound= 51 + 3(35.5)= 157.5gmol-1
0.4g/157.5gmol-1= C×50/1000
C= 2.54×10-3/0.05= 0.051M
But 1 mole of VCl3 contains 3 moles of anions
Therefore, 0.051M will contain 3×0.051M of anions= 0.153M of anions
Answer:
CH₄ will diffuse at faster rate.
Explanation:
According to Graham's law,
The rate of diffusion is inversely proportional to the square root of molar mas of gas.
It is given in question the molar mass of CH₄ is 16 g. Thus the square rot of 16 is.
√16 = 4
The molar mass of SO₂ is 64 g and the square root is,
√64 = 16
As it is stated that diffusion rate is inversely related to the square root of molar mass thus the diffusion rate of SO₂ is lower while methane diffuse faster.
The picture won’t show up
Answer:
Explanation:
Valance electrons are loosely held electrons of an atom. They are involve in chemical reaction. Consider the example of metals such as group two metals. All these have two valance electrons. They needed six electrons to complete the octet or loses two valance electrons to get complete octet. Thus its easier to remove two electrons than getting six electrons. These metals remove two electrons and form cations.
Now consider the example of nonmetals such group sixteen. They needed two electrons to get complete octet or remove six electrons to get complete octet. Thus its easier for them to get two electrons and they form anion. When group two metals cation and group sixteen anions combine they form compound and chemical reaction occur.
Group two metals also combine with halogens. Two halogens atoms combine with one alkaline earth metal atom to cancel the charge and make compound neutral.
They react with oxygen and form oxide.
2Ba + O₂ → 2BaO
2Mg + O₂ → 2MgO
2Ca + O₂ → 2CaO
Oxygen carry -2 charge while Ca, Mg and Ba +2 and make the compound neutral because charges are equal in magnitude.
With sulfur,
Mg + S → MgS
Ca + S → CaS
Ba + S → BaS
Sulfer carry -2 charge while Ca, Mg and Ba +2 and make the compound neutral because charges are equal in magnitude.
