The concentration after 225 s is 0.099 M.
As we know that, the graph of ln [A] versus time yields a straight line with slope 'k'.
So, Slope = k =
where,
k = rate constant =
t = time passed by the sample = 225 s
a = initial amount of the reactant = 0.260 M
a - x = amount left after decay process = ?
On substituting the values:
Therefore, the concentration after 225 s is 0.099 M.
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Answer : The concentration after 225 s is, 0.099 M
Explanation :
As we know that, the graph of ln [A] versus time yields a straight line with slope 'k'.
So, Slope = k =
Expression for rate law for first order kinetics is given by:
where,
k = rate constant =
t = time passed by the sample = 225 s
a = initial amount of the reactant = 0.260 M
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get
Therefore, the concentration after 225 s is, 0.099 M
0.085 moles of Al are required to form 23.6 g of AlBr₃.
Let's consider the following balanced equation for the synthesis reaction of AlBr₃.
2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)
First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).
The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885 moles of AlBr₃ are:
0.085 moles of Al are required to form 23.6 g of AlBr₃.
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Answer:
0.088 mole of Al.
Explanation:
First, we shall determine the number of mole in 23.6 g of AlBr₃.
This is illustrated below:
Mass of AlBr₃ = 23.6 g
Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol
Mole of AlBr₃ =.?
Mole = mass/Molar mass
Mole of AlBr₃ = 23.6 / 267
Mole of AlBr₃ = 0.088 mol
Next, we shall writing the balanced equation for the reaction.
This is given below:
2Al(s) + 3Br₂(l) → 2AlBr₃(s)
From the balanced equation above,
2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.
Finally, we shall determine the number of mole of Al needed for the reaction as follow:
From the balanced equation above,
2 moles of Al reacted to 2 moles AlBr₃.
Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.
(33 Points)
A.polar water, nonpolar oil
B.polar vinegar, nonpolar nail polish
C. polar bleach, polar water
D.nonpolar mineral oil, polar alcohol
Answer:
C
Explanation:
polar has unequal sharing of electrons that has the lone pairs which has the electronegativity difference. can be mixed with water.
The combination of polar bleach and polar water would likely be most miscible with each other as substances with similar molecular structures and polarities tend to mix. This is based on the rule of 'like dissolves like' in chemistry.
The question is asking which combination of substances would be most likely to mix, or be miscible, in one another. Miscibility is a property in chemistry relating to whether different substances can combine into a homogeneous mixture without separating.
The rule of thumb in chemistry is 'like dissolves like' which means that substances with similar molecular structures and polarities tend to be miscible with each other. Therefore, among the provided options, the combination of polar bleach and polar water (Option C) would be the most likely to mix with each other as they are both polar substances.
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B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4
Answer:
B. NH3 < PH3< CH4
Explanation:
Hello,
In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C, -161.5 °C and -87.7 °C, clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:
B. NH3 < PH3< CH4
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Answer: The first plastic based on a synthetic polymer was invented in 1907. Plastic sticks around in the environment for ages, threatening wildlife and spreading toxins.
Answer:
newtons 3rd law of motion
Explanation:
Answer:
pH of Buffer Solution 5.69
Explanation:
Mole of anhydrous sodium acetate =
=
= 0.18 mole
100 ml of 0.2 molar acetic acid means
= M x V
= 0.2 x 100
= 20 mmol
= 0.02 mole
Using Henderson equation to find pH of Buffer solution
pH = pKa + log
= 4.74 + log
= 4.74 + log 9
= 5.69
So pH of the Buffer solution = 5.69