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UNO [17]
4 years ago
5

For the decomposition of nitramide in aqueous solution at 25C

Chemistry
1 answer:
JulsSmile [24]4 years ago
4 0

Answer:

0.001968 M min-1

Explanation:

NH2NO2(aq) --> N2O(g) + H2O(l)

[NH2NO2],M       0.978 0.476 0.232    0.113

time, min              0          255     510      765

The average rate of disappearance over the time period form t = 0 min to t = 255 min is given as;

Average rate = Change in concentration / time taken

Change in concentration = Final Concentration - Initial concentration

Change = 0.476 - 0.978 = 0.502

time taken = 255min - 0min = 255 min

Average rate = 0.502 / 255 = 0.001968 M min-1

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Draw the structure of 4-methylcycloheptanol.
Furkat [3]

Answer:

<em>Answer is in the attachment</em>

6 0
3 years ago
What would be the new pressure if 250. сm of gas at 25.0°C and 730. mmHg pressure
IrinaVladis [17]

Answer:

2340mmHg

Explanation:

P1V1/n1T1=P2V2/n2T2

n1=n2 T1=25+271=298k

T2=300+273=573k

730×250/298=P2×150/573

P2=2340mmHg

5 0
3 years ago
The rate at which a certain Australian tree cricket chirps is 194/min at 28°C, but only 47.6/min at 5°C, From these data calcula
uysha [10]

Answer: The energy of activation for the chirping process is 283.911 kJ/mol

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at 28^0C = 194/min

K_1 = rate of reaction at  5^0C = 47.6 /min

Ea = activation energy

R = gas constant = 8.314 J/Kmol

tex]T_1[/tex] = initial temperature = 5^oC=273+5=278K

tex]T_1[/tex] = final temperature = 28^oC=273+28=301K

Now put all the given values in this formula, we get

\frac{194}{47.6}=\frac{E_a}{2.303\times 8.314}[\frac{1}{278}-\frac{1}{301}]

{E_a}=283911J/mol=283.911kJ/mol

Thus the energy of activation for the chirping process is 283.911 kJ/mol

8 0
3 years ago
Can someone please help with chemistry? Lead has a density of 10.5 g/cm^3. What is the diameter of a lead ball that has a mass o
adelina 88 [10]
Density = Mass / Volume

10.5 = 500 / x

10.5x = 500

x = 50 cm^3 (1 sig fig) volume of the sphere

Volume of a sphere = 4/3 pi r^2

pi = 3.14 r = radius

50 = 4/3 pi r^3

50•3/4 = pi r^3

37.5 = pi r^3

37.5/pi = r^3

11.9 = r^3

cube root(11.9) = r

2.3 = r

Diameter = 2•radius

Diameter = 2 • 2.3

Diameter = 5 cm (1 sig fig) diameter



3 0
3 years ago
How many hydrogen molecules are in 1.2 moles of hydrogen?
nalin [4]

Answer:

7.22 x 10²³molecules

Explanation:

Given parameters:

Number of moles of hydrogen  = 1.2moles

Unknown:

Number of molecules of hydrogen  = ?

Solution:

From the concept of moles, a mole of a substance contains the Avogadro's number of particles.

       1 mole of a substance  = 6.02 x 10²³ molecules;

So;  1.2 moles of hydrogen  = 1.2 x  6.02 x 10²³ molecules;

                                               = 7.22 x 10²³molecules

6 0
3 years ago
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