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4 years ago

What are chemical reactions?

2 answers:

5 0

2 elements or more sharing/gaining/losing to each other for a result of full outer shells of electrons of each element to become noble gases.

Bezzdna [24]4 years ago

4 0

Chemical reactions are reactions that make chemical changes to a type of something. sometimes its physically seen as well.
yw

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The name of CuO is --

Vesnalui [34]

The name is Copper(II) Oxide and/or Cupric Oxide.

Cu is Copper and O is Oxygen.

Usually when the Oxygen ion is placed after another, it's name is Oxide.

6 0

3 years ago

The table describes how some substances were formed.

lisabon 2012 [21]

Explanation:

Which is a pure substance?

1. soda

2. gasoline

3. salt water

4. carbon dioxide

carbon dioxide

Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of -7.2°C. Bromine can be classified as a

1. compound.

2. impure substance.

3. mixture.

4. pure substance.

pure substance.

6 0

3 years ago

Read 2 more answers

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

How does heat, without causing melting, damage a rock below eath’s surface

zheka24 [161]

It could be erosion

4 0

3 years ago

Question 34 (1 point)

Grace [21]

Answer:

8.33 atm

Explanation:

Xe is 5 out of (4+5) or 5 / 9 ths of the gas present

5/9 * 15 atm = 8.33 atm

4 0

2 years ago