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3 years ago

Oxidation of methane

Chemistry

1 answer:

Mandarinka [93]3 years ago

4 0

<h2><em><u>ᎪꪀsωꫀᏒ</u></em></h2>

➪Methane (CH4) is oxidized with molecu- lar oxygen (O2) to carbon dioxide (CO2).

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A gas has a volume of 13.4 L at 17C. What is the volume of the gas at standard temperature?

VashaNatasha [74]

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This law states that the volume and temperature of a gas have a direct relationship: As temperature increases, volume increases, when pressure is held constant. Heating a gas increases the kinetic energy of the particles, causing the gas to expand.

Explanation:

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3 years ago

The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport

PtichkaEL [24]

Answer:

The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport chain. The energy from the electron transport chain is used for oxidative phosphorylation.

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4 years ago

What is the molarity of 5.60 mol of sodium carbonate in 1500 ml of solution?

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Answer:

3.74 M

Explanation:

We know that molarity is moles divided by liters. The first thing to do here is convert your 1500 mL of solution to L. There's 1,000 mL in 1 L, so you need to divide 1500 by 1000:

1500 ÷ 1000 = 1.50

Now you can plug your values into the equation for molarity:

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3 years ago

What are the protons neutrons and electrons of Vanadium-52 +3 charge<br> HELP

Alex17521 [72]

Answer:

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Neutrons: 28

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Explanation:

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3 years ago

Hydrogen-3 has a half-life of 12.32 years. A sample of H-3 weighing 3.02 grams is left for 15.0 years. What will the final weigh

kaheart [24]

Answer:

$\large \boxed{\text{1.38 g}}$

Explanation:

Two important formulas in radioactive decay are

$(1) \qquad t_{\frac{1}{2}} = \dfrac{\ln 2}{k}\\\\(2) \qquad \ln \left(\dfrac{N_{0}}{N}\right) = kt$

1. Calculate the decay constant k

$\begin{array}{rcl}t_{\frac{1}{2}} &=& \dfrac{\ln 2}{k}\\\\\text{12.32 yr} &= &\dfrac{\ln 2}{k}\\\\k & = & \dfrac{\ln 2}{\text{12.32 yr}}\\\\& = & \text{0.056 26 yr}^{-1}\\\end{array}$

2. Calculate the mass remaining

$\begin{array}{rcl}\ln \left(\dfrac{A_{0}}{A}\right) &= &kt \\\\\ln \left(\dfrac{\text{3.20 g}}{A}\right) &= &\text{0.056 26 yr}^{-1}\times \text{15 yr} \\\\\ln \left(\dfrac{\text{3.20 g}}{A}\right) &= &0.8439 \\\\\dfrac{\text{3.20 g}}{A} &= &e^{0.8439} \\\\\dfrac{\text{3.20 g}}{A}&= &2.325 \\\\A &= &\dfrac{\text{3.20 g}}{2.325}\\\\&= & \textbf{1.38 g}\\\end{array}\\\text{The final mass of the sample will be $\large \boxed{\textbf{1.38 g}}$}$

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3 years ago