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3 years ago

Someone help me please

Chemistry

2 answers:

Aleonysh [2.5K]3 years ago

8 0

<u>Examples of energy:</u> Nuclear, Fossil Fuels, Solar, Wind, Hydroelectric, Geothermal, Kinetic, Mechanical, Potential, and Chemical.

<u>Non-examples:</u> love, atoms, hatred, air, matter, ideas, socket, batteries,

crimeas [40]3 years ago

6 0

<u>Examples of energy</u>: Heat, Electrical, Mechanical, Nuclear, Chemical, Kinetic and Potential.

<u>Non-Examples of energy</u>: Ideas, matter, water, paper, people and coffee.

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Consider the balanced chemical equation for the combustion of methane (CH4).

Mrac [35]

<h3>Answer:</h3>

89.6 L of O₂

<h3>Solution:</h3>

The balanced chemical equation is as,

CH₄ + 2 O₂ → CO₂ + 2 H₂O

As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂

So,

88 g CO₂ will be produced by = X L of O₂

Solving for X,

X = (88 g × 44.8 L) ÷ 44 g

X = 89.6 L of O₂

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3 years ago

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Determine the hydrogen ion concentration of solution B. [1]

bazaltina [42]

The hydrogen Ion concentration of solution B is
1.0 x 10^-5 or 0.000 010 M

You can see that this will be proportional to the amount of B's PH compared to A's

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3 years ago

What type of physical weathering is caused by the expansion of water?

skelet666 [1.2K]

Answer:

wedging

Explanation:

ice or frost wedging

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3 years ago

What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP

Dvinal [7]

Vol.250 before its to much pressure

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4 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>

8 0

3 years ago