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2 years ago

Can chlorine become sulfur

Chemistry

2 answers:

babunello [35]2 years ago

8 0

Yes sir it can. Or ma’am

aev [14]2 years ago

8 0

Yes, this is how. SCl2 is produced by the chlorination of either elemental sulfur or disulfur dichloride. The process occurs in a series of steps, some of which are: S8 + 4 Cl2 → 4 S2Cl2; ΔH = −58.2 kJ/mol. S2Cl2 + Cl2 ↔ 2 SCl2; ΔH = −40.6 kJ/mol.

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The stored form of energy based on an object's position is called....?

atroni [7]

Positional energy is the energy stored within an object due to the objects position arrangement or state positional in energy is one of the two main forms energy along with kinetic energy

6 0

3 years ago

Read 2 more answers

Arrange these molecules in order of decreasing mass: C6H14 , NO2 , Fe2O3 , and CaCO

babunello [35]

The following is the arrangement of the given molecules in decreasing order of mass: $\text{ Fe2O3} > \text{C6H14} > \text{CaCO} > \text{NO2}$

Solution:

First inorder to arrange the elements in descending order of their mass we have to calculate the molecular mass of each element. The calculation is as follows:

Mass of C6H14:

$C\rightarrow6\times12.01 = 72.06$

$H\rightarrow14\times1.008 = 14.112$

Mass of C6H14 is 86.172

Mass of NO2:

$N\rightarrow1\times14.0067 = 14.0067$

$O\rightarrow2\times15.9994 = 31.9988$

Mass of NO2 is 46.0055

Mass of Fe2O3:

$Fe\rightarrow2\times55.845 = 111.69$

$O\rightarrow3\times15.9994 = 47.9982$

Mass of Fe2O3 is 159.6882

Mass of CaCO:

$Ca\rightarrow1\times40.078 = 40.078$

$C\rightarrow1\times12.01 = 12.01$

$O\rightarrow1\times15.9994 = 15.9994$

Mass of CaCO will be 68.0874

So, the order will be $159.6882>86.172>68.0874>46.0055$ which is $\text{ Fe2O3} > \text{C6H14} > \text{CaCO} > \text{NO2}$

6 0

3 years ago

The reaction of aluminum with bromine is shown here. The equation for the reaction is

m_a_m_a [10]

Answer is (b)

2,3,1 is the stoichiometry

The values in front of the elements are the stoichiometric values
(Since Al2Br6 has no value in front, it's considered 1)

5 0

3 years ago

A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A

anygoal [31]

this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.

c1v1 = c2v2

where c1 is concentration and v1 is volume of the concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL

substituting these values in the formula

1.50 M x 50.0 mL = C x 250 mL

C = 0.300 M

concentration of the final solution is A) 0.300 M

4 0

3 years ago

Read 2 more answers

The radius of a helium atom is a 31 PM. What is the radius in nanometers?

Paladinen [302]

Answer:

Thus, the radius of the helium atom in nanometers is - 0.031 nm

Explanation:

Given that:-

The radius of the helium atom = 31 pm

Considering the conversion of length in pm to the length in nm as:-

1 pm = 0.001 nm

So,

Applying the above conversion factor in the radius of helium atom as:-

Radius = $31\times 0.001$ nm = 0.031 nm

Thus, the radius of the helium atom in nanometers is - 0.031 nm

7 0

3 years ago