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2 years ago

What is the compound of scI2

Chemistry

2 answers:

nignag [31]2 years ago

8 0

Answer: SCl2 is the chemical formula of sulfur dichloride.

Explanation:

Naily [24]2 years ago

4 0

Answer:

SCl2 is the substance formula for sulfur dichloride. This compound is made of just two components, so it is a parallel compound.

Explanation:

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The four elements that make up about 96% of body matter are ________.

AveGali [126]

An element is a pure substance that cannot be separated or divided into simpler substances by any method (physical or chemical). The four common elements in all the living organisms are carbon, oxygen, hydrogen, nitrogen.

Thus, the four elements that make up about 96 % of body matter are carbon, oxygen, hydrogen, nitrogen (c).

4 0

2 years ago

Read 2 more answers

Which aqueous solution has the highest boiling point at standard pressure?(1) 1.0 M KC1(aq) (3) 2.0 M KCl(aq)(2) 1.0 M CaC12(aq)

miss Akunina [59]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

Adding solute to water elevates the boiling point.

The elevation in boiling point (ΔTb) can be calculated using the relation:

ΔTb = i.Kb.m,

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kb is the molal elevation constant of water.

m is the molality of the solution.

(1) 1.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

(2) 2.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

(3) 1.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

(4) 2.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

So, the aqueous solution has the highest boiling point at standard pressure is: (4) 2.0 M CaCl₂(aq).

6 0

3 years ago

In a titration of 0.5 M HCl and an unknown concentration NaOH, 25.0 ml of NaOH was required to completely neutralize 25.0 ml of

baherus [9]

When an acid is neutralized by a base, that means moles of H+ = moles of OH-

moles of H+ = 0.5 M * 0.025 L HCl = 0.0125 moles H+

moles of OH- should be equal to 0.0125 moles, so
0.0125 moles = (x) * 0.025 L NaOH

x is the concentration of NaOH, which we want to find.

x = 0.5 M

The correct answer is C) 0.5 M.

6 0

3 years ago

Represent the decomposition of aluminum oxide using the same number of atoms of molecules as are

pantera1 [17]

Answer:

Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.

Explanation:

Decomposition reaction:

When a single compound break down into two or more simpler products.

For example "AB" reactant undergoes decomposition to form "A" and "B" products.

The chemical reaction is as follows.

$AB\rightarrow A+B$

The given compound is aluminium oxide.

The decomposition reaction of aluminium oxide is a follows.

$Al_{2}O_{3}\rightarrow Al+O_{2}$

The balanced equation is as follows.

$2Al_{2}O_{3}\rightarrow 4Al+3O_{2}$

Therefore, Decomposition of aluminium oxide forms aluminium atoms and oxygen atoms.

4 0

3 years ago

The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves

adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

Where ΔT is change in temperature = 0.400°C

Kf is freezing point constant of the solvent = 1.86°C/m

m is molality of the solution (Moles of solute / kg of solvent)

And i is Van't Hoff constant (1 for a nonelectrolyte)

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

6 0

2 years ago