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sergey [27]
3 years ago
12

How many moles are in 1.2x10^3 grams of ammonia, NH3?

Chemistry
1 answer:
KonstantinChe [14]3 years ago
6 0
I don't have a calculator with me right now, but that mass would be 1200 grams. Divide the given amount of grams by the molar mass of NH3, which is 17.031g/mol. (Nitrogen + 3(hydrogen)). Again, sorry I didn't have a calculator. But all you would need to do is divide 1200 by 17.031. If you need to use sig figs, your answer should have 2 because the 1.2 x 10^3 limits your amount of sig figs.
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The average kinetic energy and rms speed of N₂ molecules at STP is 5.65686 \times 10^{-21} and $493 \mathrm{~m} / \mathrm{s}$

Given,

$$\begin{aligned}&\mathrm{P}=1.013 \times 10^{5} \mathrm{~Pa} \\&\mathrm{~T}=273.15 \mathrm{~K} \\&\rho=1.25 \mathrm{~kg} / \mathrm{m}^{3}\end{aligned}$$

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K . E .=\frac{3}{2} \times $1.380649 \times 10^{-23}$ \times 273.15 K

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$$\begin{aligned}&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{P}}{\rho}} \\&\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \times 1.013 \times 10^{5}}{1.25}}=493 \mathrm{~m} / \mathrm{s}\end{aligned}$$

The average kinetic energy of a gas's particles is inversely related to its temperature. As the gas warms, the particles must travel more quickly since their mass is constant.

The average kinetic energy (K) is equal to one half of the mass (m) of each gas molecule times the RMS speed (vrms) squared.

Learn more about  average kinetic energy brainly.com/question/1599923

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C quite literally describes what electronegativity is, you can rule out most others by looking at periodic trends as attached below.

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