All categories

2 years ago

When a hydrate is heated to high temperatures, what is generally the product?

Chemistry

2 answers:

Soloha48 [4]2 years ago

8 0

Answer:

anhydrous compound

Heating a hydrate leads to an endothermic reaction that produces a residue known as the anhydrous compound. This compound is different in structure, texture and even color in some cases, from its parent hydrate.

NARA [144]2 years ago

3 0

Warming a hydrate prompts an endothermic response that creates a buildup known as the anhydrous compound. This compound is distinctive in structure, surface and even shading now and again, from its parent hydrate.

You might be interested in

What pressure is exerted by 932.3 g of CH4 in a 0.560 L steel container at 136.2 K?

Roman55 [17]

Molar mass CH4 = 16.0 g/mol

* number of moles:

932.3 / 16 => 58.26875 moles

T = 136.2 K

V = 0.560 L

P = ?

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

P x 0.560 = 58.26875 x 0.082 x 136.2

P x 0.560 = 650.76

P = 650.76 / 0.560

P = 1162.07 atm

5 0

3 years ago

List the protons, neutrons, and electrons for Ge2+ Please hurry

Nady [450]

Answer:

32, 30 and 41

Explanation:

The problem here is to find the number of:

Protons, neutrons and electrons in Ge²⁺

In this ion,

We must understand that for a net positive charge to remain on an atom, the number of protons must be greater than the number of electrons.

Ge is Germanium with atomic number of 32;

So the number of protons is 32

Since the atom has lost two electrons;

Number of electrons now is 32 - 2 = 30

Number of neutrons is 41 from the periodic table.

7 0

2 years ago

What is the pH of a 0.300 M NH₃ solution that has Kb = 1.8 × 10⁻⁵ ? The equation for the dissociation of NH₃ is: NH₃ (aq) + H₂O

nalin [4]

Answer:

11.4

Explanation:

Step 1: Given data

Concentration of the base (Cb): 0.300 M

Basic dissociation constant (Kb): 1.8 × 10⁻⁵

Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

$[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M$

Step 4: Calculate the pOH

We will use the following expression.

$pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6$

Step 5: Calculate the pH

We will use the following expression.

$pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4$

8 0

3 years ago

Three Chemistry students find a bottle of colorless liquid in the laboratory and each makes a different suggestions about the id

astra-53 [7]

Answer:

and the question is?

Explanation:

3 0

3 years ago

Two descriptions about physical quantities are given below:

gladu [14]

Answer:

Quantity A is weight and Quantity B is mass

Explanation: weight has same unit as force. Mass is the quantity of matter present in a body or object

8 0

3 years ago

Read 2 more answers