The atmosphere of earth is made of five main layers.
1) Troposphere : This is the lowest part of the atmosphere. Most of the air that makes up the atmosphere is present in this layer.
2) Stratosphere : This layer is present above troposphere and extends up to 50 km. It contains ozone layer which prevents the harmful ultraviolet rays from the sun from entering the lower troposphere
3) Mesosphere : The layer above stratosphere is known as mesosphere.
4) Thermosphere : The region lies above mesosphere.
5) Exosphere : The is the outermost region of the atmosphere.
From the above discussion we can see that the layer that lies between exosphere and mesosphere is Thermosphere
Greetings!
To find the empirical formula you need the relative atomic mass of each element!
Li = 6.9
C = 12
O = 16
You can simply change the percentages into full grams
Li = 18.8g
C = 16.3g
O = 64.9
Then you use this to find the Number of moles = amount in grams / atomic mass
Li = 18.8 ÷ 6.9 = 2.7246
C = 16.3 ÷ 12 = 1.3583
O = 64.9 ÷ 16 = 4.0562
Then divide each number of moles by the smallest value:
Li = 2.7246 ÷ 1.3583 = 2.0
C = 1.3583 ÷ 1.3583 = 1
O = 4.0562 ÷ 1.3583 = 2.9 ≈ 3
So that means that there are 2 Li, 1 C, and 3 O
Empirical formula would be:
Li₂CO₃
Hope this helps!
Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
Answer:
The critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquefied, no matter how much pressure is applied.
