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ExtremeBDS [4]
3 years ago
13

Hydrogen sulfide and concentrated sulfuric acid react to form water and sulfur dioxide as shown in the following reaction:

Chemistry
1 answer:
Law Incorporation [45]3 years ago
8 0

Answer:

2 moles of SO₂

Explanation:

Equation of reaction

H₂S + 3H₂SO₄ → 4H₂O + 4SO₂

From the equation of reaction above,

1 mole of H₂S will produce 4 moles of SO₂

1 mole of H₂S = 4 moles of SO₂

0.5moles of H₂S = zmoles of SO₂

z = (0.5*4) / 1

z = 2 moles of SO₂

0.5 moles of H₂S will produce 2 moles of SO₂

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AlladinOne [14]
I bottle:

250 * 150mg = 37500mg

If 37500mg ------- cost ------- <span> $2.95
so 1mg       ------- cost ------- x

x = 1mg*</span> $2.95 / 37500mg =  $7,87*10⁻⁵ 

II bottle

125 * 200mg = 25000mg

If 25000mg ---------- cost ---------- <span>$3.50
so 1mg       ---------- cost ---------- x

x = 1mg* </span>$3.50 / 25000mg = $0,00014=$1,4*10⁻⁴


$7,87*10⁻⁵  < $1,4*10<span>⁻⁴
</span>
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Answer:

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Explanation:

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2 years ago
Where is groundwater stored?
Tema [17]

Answer:

Answer Below! : )

Explanation:

Groundwater is stored in the tiny open spaces between rock and sand, soil, and gravel. How well loosely arranged rock (such as sand and gravel) holds water depends on the size of the rock particles.

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8 0
2 years ago
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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>

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