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iogann1982 [59]
3 years ago
9

What are prevailing winds?

Chemistry
1 answer:
Tems11 [23]3 years ago
7 0

Answer:

D) winds that blow in the same direction at a consistent speed

Explanation:

i took the quiz got it right so i know the answer please trust me i know this is right i promise with all my heart

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What does the Periodic Table show?
Mamont248 [21]
Atomic number
atomic mass
group|family
periods
and element symbols
8 0
3 years ago
Mass of 0.432 moles of C8H9O4?
Fiesta28 [93]
Find one mole
8 C = 8 * 12 = 96
9 H = 1 * 9 = 9
4 O = 4 *16 = 64
Total = 169 

1 mol = 169 grams.
0.432 mol = x

1/0.432 = 169/x
x = 0.432 * 169
x = 73.0 grams
7 0
3 years ago
Read 2 more answers
What is the effect of adding more CO2 to the following equilibrium reaction? CO2 + H2O H2CO3 A. More H2CO3 is produced. B. More
Gre4nikov [31]
I think the answer is D no change. Though you add more CO2, but the pressure is not mentioned. If the pressure is constant and the reaction is already balanced, the H2O is also saturation and can not absorb more CO2.
6 0
2 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
Science question:<br> Could you have seasons without<br> revolution? Why or why not?
marusya05 [52]

Answer:

No

Explanation:

The Earth would not have seasons if there is no revolution because the temperatures would not change.

4 0
2 years ago
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