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Snezhnost [94]
4 years ago
13

When hydrogen is burned in oxygen to form water, the composition of water formed does not depend on the amount of oxygen reacted

. interpret this in terms of the law of definite proportion?
Chemistry
2 answers:
VLD [36.1K]4 years ago
4 0
Law of Definite Proportions states that a compound is composed of the same ratio of elements present. The formation of water has a reaction:

H2 + 1/2 O2 = H2O

The limiting reactant would be the element with the least ratio of molar mass to stoichiometric coefficient, and that would be H2, not O2.
Vedmedyk [2.9K]4 years ago
4 0

Answer:  

Explanation:  Law of Definite Proportions state that in a chemical compound, the constituent element will always exist in a fixed ratio which does not depend on the source of preparation.

So when hydrogen is burned in oxygen to form water, the composition of water formed does not depend on the oxygen but it depends on the hydrogen itself.

H_{2}O + 1/2 O_{2} \rightarrow H_{2}O

Thus the reaction usually depends on the limiting reagent ad hence , in the above reaction, hydrogen is the limiting reagent. Thus the reaction will depend on the hydrogen, not on oxygen.

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What are the physical properties we use to classify metals, nonmetals, and metalloids? *
laila [671]

Answer:

Biden 2020

Explanation:

7 0
3 years ago
A typical double bond __________________________. a. is stronger and shorter than a single bond b. consists of one σ bond and on
leva [86]

Answer:

All of the above answers are correct.

Explanation:

Double bonds are stronger that single bonds because they have both sigma and pi bonds whereas single bonds contain only sigma bond. They also involve the sharing of electron pairs and heavily imparts rigidity to a molecule.

5 0
3 years ago
The density of ethanol is 0.789 g/mL. If you need 460.0 g of ethanol for an experiment, how many milliliters would you measure o
4vir4ik [10]

Answer:

The volume is 583.02 mL

Explanation:

Given that the density is:-

\rho=0.789\ g/mL

The mass, m=460.0\ g

The expression for the calculation of density is shown below as:-

\rho=\frac{m}{V}

Using the above expression to calculate the volume as:-

V=\frac{m}{\rho}

Applying the values to calculate the volume as:-

V=\frac{460.0\ g}{0.789\ g/mL}=583.02\ mL

<u>The volume is 583.02 mL.</u>

3 0
4 years ago
A scientist measures the standard enthalpy change for the following reaction to be -139.5 kj :
jeyben [28]

Answer:

A scientist measures the standard enthalpy change for the following reaction to be -139.5 kj :

h2(g) + c2h4(g)c2h6(g)

based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of c2h4(g) is _____ kj/mol

Explanation:

Hydrogen ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K)

H2 (g)

0

0

130.7

Carbon ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K)

C2H6 (g)

-84.7

-32.8

229.6

7 0
2 years ago
Determine the bonding type for boron trihydride given the electronegativity on the Pauling scale for boron is 2.0 and for hydrog
Anika [276]

Answer: The bond between boron and hydrogen in boron trihydride is covalent bond.

Explanation:

The type of bonding between the atoms forming a compound is determined by using the electronegativity difference between the atoms. According to the pauling's electronegativity rule:

  • If \Delta \chi=0, then the bond is non-polar.
  • If \Delta \chi\leq 1.7, then the bond will be covalent.
  • If \Delta \chi>1.7, then the bond will be ionic.

We are given:

Electronegativity for boron = 2.0

Electronegativity for hydrogen = 2.1

\Delta \chi=\chi_{H}-\chi_{B}\\\\\Delta \chi=2.1-2.0=0.1

As, \Delta \chi is less than 1.7 and not equal to 0. Hence, the bond between boron and hydrogen is covalent bond.

3 0
3 years ago
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