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VladimirAG [237]
3 years ago
10

The decomposition of mercury (ii) oxide at high temperature, is it an endothermic or exothermic process? Write a chemical reacti

on for this process.

Chemistry
2 answers:
gulaghasi [49]3 years ago
6 0

Answer:

that will be endothermic reaction...... as oxides of mercury decomposes and break s into simpler elements by absorbing energy

Explanation:

hope it helped u buddy

algol133 years ago
3 0

Answer: endothermic, 2HgO\rightarrow 2Hg+O_2

Explanation:

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

All decomposition reactions are endothermic as heat is absorbed by the reactants to break the bonds in the reactants.

Decomposition of HgO is given by:

2HgO\rightarrow 2Hg+O_2

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How many molecules are in 1.5 moles of carbon dioxide
Karo-lina-s [1.5K]

Answer:

<h2>9.03 × 10²³ molecules</h2>

Explanation:

The number of molecules can be found by using the formula

N = n × L

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 1.5 × 6.02 × 10²³

We have the final answer as

<h3>9.03 × 10²³ molecules</h3>

Hope this helps you

3 0
3 years ago
if .40 L if water is added to the volume of cup 3, what would be the new molarity of a 2 M solution of kool-aid
stiks02 [169]

Answer:

0.7692 M ≅ 0.77 M.

Explanation:

  • It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.
  • It can be expressed as:

<em>(MV) before dilution = (MV) after dilution.</em>

M before dilution = 2.0 M, V before dilution = 0.25 L.

M after dilution = ??? M, V after dilution = 0.25 L + 0.40 L = 0.65 L.

∴<em> M after dilution = (MV) before dilution/(V) after dilution</em> = (2.0 M)(0.25 L)/(0.65 L) =<em> 0.7692 M ≅ 0.77 M.</em>

8 0
3 years ago
What type of mixture will not allow light to pass through
igor_vitrenko [27]
Suspension.  The particles are big enough for the eye to see, and will separate if left sitting.
5 0
4 years ago
Consider the solubilities of a particular solute at two different temperatures. Temperature ( ∘ C ) Solubility ( g / 100 g H 2 O
grin007 [14]

Answer:

21.28 grams solute can be added if the temperature is increased to 30.0°C.

Explanation:

Solubility of solute at 20°C = 32.2 g/100 grams of water

Solute soluble in 1 gram of water = \frac{32.2}{100}g=0.322 g

Mass of solute in soluble in 56.0 grams of water:

0.322\times 56.0=18.032 g

Solubility of solute at 30°C = 70.2g/100 grams of water

Solute soluble in 1 gram of water = \frac{70.2}{100}g=0.702 g

Mass of solute in soluble in 56.0 grams of water:

0.702 \times 56.0=39.312 g

If the temperature of saturated solution of this solute using 56.0 g of water at 20.0 °C raised to 30.0°C

Mass of solute in soluble in 56.0 grams of water 20.0°C = 18.032 g

Mass of solute in soluble in 56.0 grams of water at 30.0°C = 39.312 g

Mass of of solute added If the temperature of the saturated solution increased to 30.0°C:

39.312 g - 18.032 g = 21.28 g

21.28 grams solute can be added if the temperature is increased to 30.0°C.

8 0
3 years ago
True or false , the viscosity of water is much greater than the viscosity ofhoney.
a_sh-v [17]
False... Honey is greater in velosity then water
4 0
4 years ago
Read 2 more answers
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