This is an incomplete question, here is a complete question.
If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.
The reaction
It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.
Answer : The concentration of after 9.0 seconds is, 0.00734 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 0.80 M⁻¹s⁻¹
t = time taken = 142 second
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.0440 M
Putting values in above equation, we get:
![0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)](https://tex.z-dn.net/?f=0.80M%5E%7B-1%7Ds%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7B142s%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.0440M%29%7D%5Cright%29)
![[A]=0.00734M](https://tex.z-dn.net/?f=%5BA%5D%3D0.00734M)
Hence, the concentration of after 9.0 seconds is, 0.00734 M
Answer:
2
Explanation:
Mass of water molecule = mass of hydrated salt - mass of anhydrous salt
Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.
Number of moles = mass / molarmass
Molar mass of water = 18.015g/mol
No. of moles of water = 0.74 / 18.015 = 0.0411 moles.
Mass of BaCl2 present =?
1 mole of BaCl2 = 208.23 g
X mole of BaCl2 = 4.26 g
X = (4.26 * 1) / 208.23
X = 0.020
0.020 moles is present in 4.26g of BaCl2
Mole ratio between water and BaCl2 =
0.0411 / 0.020 = 2
Therefore 2 molecules of water is present the hydrated salt.
Answer:
a. The original temperature of the gas is 2743K.
b. 20atm.
Explanation:
a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:
T₁n₁ = T₂n₂
<em>Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.</em>
<em />
<em>Replacing with values of the problem:</em>
T₁n₁ = T₂n₂
X*7.1g = (X+300)*6.4g
7.1X = 6.4X + 1920
0.7X = 1920
X = 2743K
<h3>The original temperature of the gas is 2743K</h3><h3 />
b. Using general gas law:
PV = nRT
<em>Where P is pressure (Our unknown)</em>
<em>V is volume = 2.24L</em>
<em>n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)</em>
R is gas constant = 0.082atmL/molK
And T is absolute temperature (2743K)
P*2.24L = 0.20mol*0.082atmL/molK*2743K
<h3>P = 20atm</h3>
<em />
Explanation:
The - 3 degree C( carbon atom) 2p atomic orbital + methyl C-H sigma molecular orbital because one C-H bond has to dissolve its bond and provide the H that is sigma molecular orbital and the carbonation is type 3 degree sp2 carbon.
Hyperconjugation is the stabilizing effect arising from the electrons ' engagement in a π-bond (usually C-H or C-C) with a neighboring empty or partly filled p-orbital or π-orbital to provide an expanded molecular orbital that enhances system stability.
Given mass of KNO₃=346g
Molar mass of KNO₃=(39.098)+(14)+(15.99*3)=101.068gmol⁻¹
Volume of Solution=750ml=0.75dm³
Molarity=(mass of solute/molar mass of solute)*(1/volume of sol. in dm³)
=(346/101.068)*(1/0.75)
=4.56 mol dm⁻³
