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olya-2409 [2.1K]
3 years ago
11

Predict which of the substances, NH3, N2, CH2Cl2, Cl2, CCl4 has

Chemistry
1 answer:
gulaghasi [49]3 years ago
6 0

Answer:

djsjwjwjdjxjdbs djwuqidx8diw d xbd

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Complete each equation with a number that is true.<br>5 * _ =15
natali 33 [55]

The answer is 3. As 5 * 3 = 15.

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5 0
3 years ago
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A gas cylinder of volume 5.00 l contains 1.00 g of ar and 0.500 g of ne. the temperature is 275 k. find the partial pressure of
amid [387]
<span>11.3 kPa The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = Absolute temperature We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon. Atomic weight argon = 39.948 Atomic weight neon = 20.1797 Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol Now take the ideal gas equation and solve for P, then substitute known values and solve. PV = nRT P = nRT/V P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L P = 113.8892033 L*kPa / 5.00 L P = 22.77784066 kPa Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles. 0.024777375 mol / 0.049809918 mol = 0.497438592 Now multiply by the pressure 0.497438592 * 22.77784066 kPa = 11.33057699 kPa Round the result to 3 significant figures, giving 11.3 kPa</span>
8 0
3 years ago
Consider the solubilities of a particular solute at two different temperatures.
GenaCL600 [577]

Explanation:

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4 0
2 years ago
A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is th
Kay [80]

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

0.0800 mol LiCl + 4.444 mol = mol LiCl

mol LiCl - 0.0800 mol LiCl = 4.444 mol

0.92 mol LiCl = 4.444 mol

mol LiCl = 4.83 m

7 0
3 years ago
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