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2 years ago

Why is a hydrogen bond a relatively weak bond?

1 answer:

5 0

Hydrogen bonds are intermolecular (“between-molecule”) bonds, rather than intramolecular (“within-molecule”) bonds. They occur not only in water but in other polar molecules in which positive hydrogen atoms are attracted to negative atoms in nearby molecules. Hydrogen bonds are relatively weaker as chemical bonds. For example, they are much weaker than the bonds holding atoms together within molecules of covalent compounds.

Hydrogen bond is weaker than ionic and covalent bonds. Example: Water molecules are held to each other by intermolecular forces of attraction. Covalent bonds are the strongest bonds, meaning that atoms are actually held together by the physical sharing of electrons.

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What do the coefficients in the following balanced chemical equation mean?

mafiozo [28]

It's either B or D because there would be 2 coefficients of hydrogen and one of oxygen

6 0

3 years ago

Lead (II) nitrate reacts with ammonium carbonate to produce ammonium nitrate

vitfil [10]

Pb(NO₃)₂ + (NH₄)₂CO₃ → PbCO₃ + 2 NH₄NO₃

Explanation:

Reaction of lead (II) nitrate with ammonium carbonate will produce lead (II) carbonate and ammonium nitrate.

The balanced chemical equation is:

Pb(NO₃)₂ + (NH₄)₂CO₃ → PbCO₃ + 2 NH₄NO₃

To balance the chemical equation the number of atoms of each element entering the reaction have to be equal to the number of atoms of each element leaving the reaction, in order to conserve the mass.

Learn more about:

balancing chemical reactions

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#learnwithBrainly

3 0

3 years ago

Determine the empirical formula of a compound containing 48.38 grams of carbon, 6.74 grams of hydrogen, and 53.5 grams of oxygen

Darya [45]

Answer: The empirical formula of the compound becomes $CH_2O$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

We are given:

Mass of C = 48.38 g

Mass of H = 6.74 g

Mass of O = 53.5 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of C}=\frac{48.38g}{12g/mol}=3.023 mol$

$\text{Moles of H}=\frac{6.74g}{1g/mol}=6.74 mol$

$\text{Moles of O}=\frac{53.5g}{1g/mol}=3.34 mol$

Step 2: Calculating the mole ratio of the given elements.

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 3.023 moles

$\text{Mole fraction of C}=\frac{3.023}{3.023}=1$

$\text{Mole fraction of H}=\frac{6.74}{3.023}=2.23\approx 2$

$\text{Mole fraction of O}=\frac{3.34}{3.023}=1.105\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 2 : 1

Hence, the empirical formula of the compound becomes $CH_2O$

5 0

2 years ago

What's the answer to the question in the picture??

Burka [1]

Answer:

Explanation:

3 0

3 years ago

The empirical formula for trichloroisocyanuric acid, the active ingredient in many household bleaches is OCNCI. The molar mass o

TEA [102]

Answer:

These two are equivalent and valid:

$C_3Cl_3N_3O_3$

$Cl_3(CN)_3O_3$

Explanation:

The molecular superscripts for each atom in the molecular formula are determined by the number of times that the mass of the empirical formula is contained in the molar mass.

1. Determine the mass of the empirical formula:

$OCNCl$ :

Atomic masses:

O: 15.999g/mol
C: 12.011g/mol
N: 14.007g/mol
Cl: 35.453g/mol

Total mass:

15.999g/mol + 12.011g/mol + 14.007g/mol + 35.453g/mol = 77.470g/mol

2. Divide the molar mass by the mass of the empirical formula:

232.41g/mol / 77.470g/mol = 3

3. Multiply each superscript of the empirical formula by the previous quotient: 3

$O_3C_3N__3Cl_3$

Or:

$C_3Cl_3N_3O_3$

You might also write CN as a group:

$Cl_3(CN)_3O_3$

8 0

3 years ago