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3 years ago

14

A cube of iron at 20C is placed in contact with a cube of copper at 60C. Which statement describes the initial flow of heat betw

een the cubes?

1 answer:

Anestetic [448]3 years ago

7 0

Answer:

the iron is more condensed then the copper more electricity

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What is the concentration of 60 mL of H3PO4 if it is neutralized by 225 mL of 2 M Ba(OH)2?

goldfiish [28.3K]

7.5 M is the concentration of 60 ml of H3PO4 if it is neutralized by 225 ml of 2 M Ba(OH)2.

Explanation:

Data given:

volume of phosphoric acid, Vacid =60 ml

volume of barium hydroxide, Vbase = 225 ml

molarity of barium hydroxide, Mbase = 2M

Molarity of phosphoric acid, Macid =?

the formula for titration is used as:

Macid x Vacid = Mbase x Vbase

rearranging the equation to get Macid

Macid = $\frac{Mbase x Vbase}{Vacid}$

Macid = $\frac{225 X 2}{60}$

Macid = 7.5 M

the concentration of the phosphoric acid is 7.5 M and the volume is 60 ml. Thus 7.5 M solution of phosphoric acid is used to neutralize the barium hydroxide solution of 2M.

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3 years ago

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Help me i mark brain thing

o-na [289]

Answer:

Becuase they can go through mountains and hills, they curve of of those.

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2 years ago

The difference between the boiling point of a pure solvent and the boiling point of a solution of a nonelectrolyte in the same s

Sati [7]

Answer: A) Boiling-Point Elevation

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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 $i = \dfrac{\Delta T_f}{k_f \times m}$

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

$\mathbf {i = 1.51 }$

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3 years ago

Compute 9.3456+2140.56. Round the answer appropriately.

vichka [17]

This is just addition. Put 2140.56 on top, line up 9.3456 under it appropriately. Doing this will give you the answer: 2149.9056

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3 years ago