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Olegator [25]
3 years ago
14

Which element will most easily lose an electron?

Chemistry
2 answers:
Vadim26 [7]3 years ago
7 0

Answer:

C

Explanation:

Group 1 elements lose an electron to form charged positive ions.

Potassium is from group 1.

Potassium is most likely to lose an electron.

tresset_1 [31]3 years ago
5 0

Answer:

C

Explanation:

The elements that will most likely lose an electron are those that are in Group 1. This is because they only have one valence electron in their outer shell. Atoms want to have 8 electrons in their outer shell and in order for the elements of Group 1 to do so, they could either lose one electron or gain 7 but losing one is easier to do. Out of these answer choices, the only one that belongs to Group 1 is potassium.

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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
What volume will 1.27 moles of helium gas occupy at 80.00 °C and 1.00 atm?
White raven [17]

Answer:

36.8 L

Explanation:

We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 80 °C

T(K) = 80 + 273

T(K) = 353 K

Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:

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Volume (V) =?

PV = nRT

1 × V = 1.27 × 0.0821 × 353

V = 36.8 L

Thus, the volume occupied by the helium gas is 36.8 L

5 0
3 years ago
What is the mole fraction of methanol in a solution that contains 6.0 mol of methanol and 3.0 mol of water? The formula for meth
Ivenika [448]
The mole fraction of a product is the number of moles of the product divided by the total number of moles of the solution.

Here moles of methanol = 6.0 moles

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Mole fraction of methanol = 6.0 / 9.0 = 0.67

Answer: 0.67  
4 0
3 years ago
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