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3 years ago

List four observations that suggest chemical change is occurring

Chemistry

1 answer:

allsm [11]3 years ago

8 0

Answer:

Examples of Chemical Changes

Burning wood.

Souring milk.

Mixing acid and base.

Digesting food.

Cooking an egg.

Heating sugar to form caramel.

Baking a cake.

Rusting of iron.

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Classify each of the following as energy primarily transferred as HEAT or energy primarily transferred as WORK.

IgorLugansk [536]

number 1 is work number 2 is heat

6 0

3 years ago

A metal pellet with a mass of 100.0 g, originally at 116°C, is dropped into a cup of water, initially at

Alina [70]

Answer:

C, 42g

Explanation:

In thermal equilibrium, both bodies (metal pellet and water) both have the same final temperature (46.3°C).

Assuming no heat is lost to surroundings,

the energy lost from metal pellet = energy gained for water

Since E = mc∆T

(energy = mass x specific heat capacity x temperature change)

mc∆T (metal pellet) = mc∆T (water)

100 x 0.568 x (116-46.3) = m 4.184 (46.3 - 23.8)

3958.96 = 94.14m

m = 42g

6 0

3 years ago

A neutral solution has a pH =

Aleonysh [2.5K]

Answer:

Explanation:

A neutral solution has a pH=7.

A basic solution has a pH>7.

An acidic solution has a pH<7.

3 0

3 years ago

Which statement below is not true?

Verizon [17]

Hi I believe the answer is B
Hope this helps you

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

3 years ago