Answer:
1 dihydrogen reacts with 1 chlorine to form two hydrogen chloride (edit: hydrogen chloride is also referred to as hydrochloric acid)
The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.
<h3>What is an empirical formula?</h3>
It is the minimum ratio between the elements that form a compound.
- Step 1: Divide each percentage by the molar mass of the element.
C: 18.1/12.01 = 1.51
H: 2.27/1.01 = 2.25
Cl: 79.8/35.45 = 2.25
- Step 2: Divide all the numbers by the smallest one.
C; 1.51/1.51 = 1
H: 2.25/1.51 ≈ 1.5
Cl: 2.25/1.51 ≈ 1.5
- Step 3: Multiply all the numbers by 2 so all of them are whole.
C: 1 × 2 = 2
H: 1.5 × 2 = 3
Cl: 1.5 × 2 = 3
The empirical formula is C₂H₃Cl₃.
The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.
Learn more about empirical formula here: brainly.com/question/1603500
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Answer:
kinetic energy
Explanation:
When the object is released, the gravitational potential energy is gradually converted into kinetic energy as it picks up speed.
Answer:
Explanation:
So, the formula for the compound should be:
Now we assume that we have 1 mol of substance, so we can make calculations to know the molar mass of element X, as follows:
So we have that 6 moles weight 212.7g, and we can make a rule of three to know the weight of compound X:
As we used 1 mol, we know that the molar mass is 32.06g/mol
So the element has a molar mass of 32.06 g/mol and an oxidation state of +6, with this information, we can assure that the element X is sulfur, so the compound is 
Answer:
Explanation:
This is a limiting reactant problem.
Mg(s)
+
2HCl(aq)
→
MgCl
2
(
aq
)
+ H
2
(
g
)
Determine Moles of Magnesium
Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).
4.86
g Mg
×
1
mol Mg
24.3050
g Mg
=
0.200 mol Mg
Determine Moles of 2M Hydrochloric Acid
Convert
100 cm
3
to
100 mL
and then to
0.1 L
.
1 dm
3
=
1 L
Convert
2.00 mol/dm
3
to
2.00 mol/L
Multiply
0.1
L
times
2.00 mol/L
.
100
cm
3
×
1
mL
1
cm
3
×
1
L
1000
mL
=
0.1 L HCl
2.00 mol/dm
3
=
2.00 mol/L
0.1
L
×
2.00
mol
1
L
=
0.200 mol HCl
Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,
2.01588 g/mol
0.200
mol Mg
×
1
mol H
2
1
mol Mg
×
2.01588
g H
2
1
mol H
2
=
0.403 g H
2
0.200
mol HCl
×
1
mol H
2
2
mol HCl
×
2.01588
g H
2
1
mol H
2
=
0.202 g H
2
The limiting reactant is
HCl
, which will produce
0.202 g H
2
under the stated conditions.
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