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Norma-Jean [14]
3 years ago
11

What makes up 90 percent of interstellar gas? hints what makes up 90 percent of interstellar gas? hydrogen carbon dioxide helium

carbon monoxide?
Chemistry
1 answer:
masha68 [24]3 years ago
7 0
Despite being an extremely rare gas on Earth, hydrogen is the most common element in the universe. This is due to the fact that hydrogen is the fuel for stars, as it undergoes the fusion nuclear reaction to form helium and release the energy to sustain the star. Therefore, the answer to this question is hydrogen.
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Which of the following is the conjugate acid of clo?
pshichka [43]

Explanation:

B. HCIO these is the answer

4 0
3 years ago
A certain liquid X has a normal boiling point of 133.60°C and a boiling point elevation constant Kb= 2.46°C kg mol^-1.Calculate
Afina-wow [57]

Answer:

136.63 °C

Explanation:

ΔTb=Tb solution - Tb pure

Where; Tb pure = 133.60°C

molar mass of solute = 121.14 g/mol

number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles

molality = 0.431 moles/350 * 10^-3 = 1.23 molal

Then;

ΔTb = Kb * m * i

Kb = 2.46°C kg mol^-1

m = 1.23 molal

i = 1

ΔTb = 2.46 * 1.23 * 1

ΔTb = 3.03 °C

Hence;

Tb solution = ΔTb + Tb pure

Tb solution = 3.03 °C + 133.60°C

Tb solution = 136.63 °C

6 0
3 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
How can wind produce erosion?
Virty [35]

Answer:

A is the coorrecct answuer

Explanation:

8 0
3 years ago
Please help me. Quickly!!!
Tomtit [17]

Answer:

500

Explanation:

8 0
3 years ago
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