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velikii [3]
1 year ago
5

Why are the properties of a compound different than the elements that make it up.

Chemistry
1 answer:
nexus9112 [7]1 year ago
3 0

The properties of compounds are different, and in some cases very different, from those of the elements that compose them. This is because elements combine in compounds to become completely different substances with unique properties.

Compounds contain atoms of different elements chemically bonded in fixed ratios. An element is a pure chemical substance made up of atoms of the same kind. Compounds contain different elements in fixed ratios, arranged in a defined way by chemical bonds.

This is caused by different base properties where each element has a different number. Due to the number of electrons and differences in shell and valence electrons, different elements bond differently to certain other elements.

Compounds:

is composed of atoms of two or more different elements bonded together, and

can be broken down into simpler types of matter (elements) by chemical (rather than physical) means. from its components, and.

always contains the same proportions of atomic components.

Learn more about elements combine here;

brainly.com/question/6258301

#SPJ4

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Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 6.9 g
satela [25.4K]

Answer:

5.6gNa_2SO_4

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4

Best regards.

5 0
3 years ago
Which of the following changes
Alisiya [41]
Exothermic reactions
5 0
3 years ago
An unknown metal absorbs 61 J of heat, and its temperature increases by 29∘C. What is the heat capacity of the metal?
aniked [119]

Answer:

2.103 J/C

Explanation:

Quantity of heat = Heat Capacity * Temperature change

Heat Capacity = Quantity of heat / Temperature Change

Heat Capacity = 61/29

Heat Capacity = 2.103 J/C

3 0
3 years ago
Complete the following single replacement reaction. If they don’t react, just write “NR”
Kipish [7]

Here we have to complete the given single replacement reactions.

The replacement reactions are-

1) Fe (s) + CuCl₂ (aq) → FeCl₂ (aq) + Cu (s)

2) Cu (s) + FeCl₂ (aq) → NA

3) K (s) + NiBr₂ (aq) → NA

4) Ni (s) + KBr (aq) → NiBr₂ (aq) + K (s)

5) Zn (s) + Ca(NO₃)₂ (aq) → Zn(NO₃)₂ (aq)  + Ca (s)

6) Ca (s) + Zn(NO₃)₂ (aq) → NA

The replacement reactions can be explained in light of the redox potential.

The standard reduction potential of the half cells involved in these reactions are:

Fe²⁺ + 2e → Fe (E° = -0.441V); Cu²⁺ + 2e → Cu (E° = 0.674V)

Ni²⁺ + 2e → Ni (E° = -0.23V); Zn²⁺ + 2e → Zn (E° = -0.763V)

We know the half cell reactions in which the standard reduction potentials are positive are allowed.

1) The reaction is possible as Cu²⁺/Cu and Fe/Fe²⁺ standard reduction potentials are positive.

2) The reaction is not possible as Cu/Cu²⁺ and Fe²⁺/Fe standard reduction potentials are negative.

3) The reaction is not possible as Ni²⁺/Ni standard reduction potential is negative.

4) The reaction is possible as Ni/Ni²⁺ standard reduction potential is positive.

5) The reaction is possible as Zn/Zn²⁺ standard reduction potential is positive.

6) The reaction is possible as Zn²⁺/Zn standard reduction potential is negative.

4 0
3 years ago
If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?
Alborosie

Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L  - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

new volume -  125 mL + 25 mL = 150 mL

therefore new molarity is

c = 0.01875 mol / 0.150 L  = 0.125 M

answer is 0.125 M

7 0
3 years ago
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