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Over [174]
3 years ago
5

Match the functions with the corresponding cell structure.

Chemistry
2 answers:
goldenfox [79]3 years ago
6 0

Answer:

1.C

2.A

3. B

Explanation:

mezya [45]3 years ago
5 0

Answer:

1c 2a 3b

Explanation:

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The last element in any period always has:
brilliants [131]

Answer: The correct option is The properties of a noble gas.

Explanation: There are 7 periods in the periodic table.

The last element of each period are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) and Ununoctium (Uuo).

  • The electronic configuration for Helium is 1s^2. For He, The outermost electrons are 2.
  • The electronic configuration for all the other elements is ns^2ns^6 ( where, n = 2, 3, 4, 5, 6 and 7 respectively). For all the other gases, the outermost electrons are 8.

All these elements have stable electronic configuration and are not reactive in nature. Hence, they are considered as noble gases.

Therefore, the last element of each period always have the properties of a noble gas.

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3 years ago
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What is the PH of a 7.5x10^-3 M OH solution
Triss [41]
pH: 11.8750612634
pOH: 2.12493873661
[H+]: 1.33333333333E-12
[OH-]: 0.0075
BASE
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3 years ago
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Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
What is wrong with this graph? *<br> 1 point<br> 7<br> 6-<br> 2<br> 0
snow_lady [41]

Answer:

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Explanation:

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