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kirill115 [55]
3 years ago
14

Solve the problem. Express your answer to the correct number of significant figures

Chemistry
1 answer:
Shalnov [3]3 years ago
4 0

Answer:

Explanation:

30.53131

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Which statement is true? A molecule having a covalent bond can be ionic. A molecule having a covalent bond is always polar. A mo
Marysya12 [62]
The statement that is true is that A molecule having a covalent bond may be polar or Nonpolar.
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Name an element which matches the description
ikadub [295]

1) Uranium

2) Carbon in the form of graphite

3) Palladium

Explanation:

1) In the nuclear reactor the fuel is represented by some uranium isotopes which are easily fissionable.

2) Graphite is made from carbon sheets in which there are free electrons that are moving freely so the carbon, in this form, is able to conduct electricity.

3) Commonly inert electrodes are made from platinum (Pt) or gold (Au) metals, because of the propriety of this metals to not interact chemically very easily with various substances.

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uranium isotopes

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graphite

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3 years ago
What is the name for s2o7
Arada [10]

Answer:

Disulfate Ion {2-} Name: Disulfate Ion {2-} Formula: S2O7.

Explanation:

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3 years ago
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What ion depolarizes the membrane when it diffuses into the axon of a neuron?
Agata [3.3K]

Answer:Sodium ions

Explanation:

The membrane potential of most cells is negative, when sodium ions enters the cells, it can reverse the polarity and makes it positively charge.

7 0
3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
2 years ago
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