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Charra [1.4K]
3 years ago
10

if a fast moving cold air mass overtakes a slow-moving warm air mass in July, what type of weather would you predict to occur?

Chemistry
1 answer:
notka56 [123]3 years ago
6 0
After it passes an area there will be colder<span>, drier </span>air<span> and clear skies. This </span>occurs when a fast-moving warm air mass overtakes a slowly moving cold air mass<span>.</span>
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What is the theoretical yield of cyclohexene (in grams) that could be formed from 0.105 moles of cyclohexanol and 0.0444 moles o
zubka84 [21]

Answer:

3.65 g of cyclohexene

Explanation:

Cyclohexanol + phosphoric acid ----> cyclohexene

The reaction is 1:1 hence the limiting reactant is phosphoric acid.

Hence,

1 mole of phosphoric acid yields 1 mole of cyclohexene

0.0444 moles of phosphoric acid yields 0.0444 moles of cyclohexene

Theoretical yield = number of moles of cyclohexene × molar mass of cyclohexene

Theoretical yield = 0.0444 moles of cyclohexene × 82.143 g/mol

Theoretical yield = 3.65 g of cyclohexene

6 0
3 years ago
What would happen to the following endothermic reaction that is in equilibrium if heat is added? N2O4 (g) Two arrows stacked on
Ber [7]
The reaction N2O4 (g) <--> 2NO2 (g) is endothermic, meaning that it consumes heat to move towards formation of the products.
According to Le Chatelier's Principle, therefore, if heat is added, more product (NO2) will be produced, and equilibrium would shift towards the right side. This is choice 3.
6 0
3 years ago
Read 2 more answers
1. A 25 g rock is placed in a graduated cylinder with water. The volume of the liquid rises from 18.3 mL to 21.4 mL Calculate th
MaRussiya [10]

Answer:

1. \rho=8.06g/cm^3

2. H=10cm

Explanation:

Hello,

1. In this case, since the volume of the rock is obtained via the difference between the volume of the cylinder with the water and the rock and the volume of the cylinder with the water only:

V=21.4mL-18.3mL=3.1mL

Thus, the density turns out:

\rho=\frac{m}{V}=\frac{25g}{3.1cm^3}  \\\\\rho=8.06g/cm^3

2. In this case, given the density and mass of aluminum we can compute its volume as follows:

V=\frac{m}{\rho}=\frac{1080g}{2.7g/cm^3}=400cm^3

Moreover, as the volume is also defined in terms of width, height and length:

V=W*H*L

The height is computed to be:

H=\frac{V}{L*W}=\frac{400cm^3}{5cm*8cm}\\ \\H=10cm

Best regards.

3 0
3 years ago
if atoms of lithium and sodium easily give up a single electron which other elements also have this property
Mars2501 [29]
Basically all of the elements found in Group I of the periodic table also have this property. The ability to easily give up a single valence electron.
6 0
3 years ago
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
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