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Len [333]
3 years ago
12

Anna stated that ionic compounds have high melting points and low boiling points. She said that this demonstrates that ionic com

pounds have strong bonds because a lot of energy is needed to break the electrical forces that hold the bonds together. Which statement best describes Anna’s error?
Chemistry
2 answers:
liubo4ka [24]3 years ago
8 0

Answer: Anna stated that ionic compounds have high melting point and low boiling point. The error in the statement is that ionic compound have low boiling point, instead ionic compounds have high boiling point, because in an ionic compound, the force of attraction working between two ions is very strong and hence the bonds present are very strong, and a lot of energy is needed to break them

Eduardwww [97]3 years ago
6 0

Answer:

c

Explanation:

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Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39
nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1)   Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)    ΔH = −23 kJ

2)   3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)   ΔH = −39 kJ

3)   Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3:  Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

<u>This equation we add to the second equation</u>

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  = 2*18 + (-39) = 36 - 39 =  -3 kJ

<u>This new equation we will divide by 3 </u>

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

<u>Now we will substract this new equation from the first equation</u>

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

<u>The next equation we will divide by 2</u>

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0
3 years ago
What is the ionization charge for the elements found in Group II?
ser-zykov [4K]

Answer: a + 2

Explanation: Alkali Earths or Group II has an ionization charge of a + 2. Alkali Metals have a ionization a + 1. Halogens or cold elements have a ionization of a +3.

5 0
3 years ago
How many atoms are in this given equation 6Mn3C2
Ilia_Sergeevich [38]

Answer:

There are three atoms in the equation

6 0
3 years ago
If, in a peptide chain, there were 85 amino acids each joined by peptide bonds, how many n-terminus groups would be present?
Crank
If, in a peptide chain, there were 85 amino acids each joined by peptide bonds, there would only be 1 N-terminus group that would be present. The N-terminus group is always the start of the chain of a amino acid chain or a protein or a polypeptide. It refers to the free amine group present that is located at the end part of the chain. So, that no matter how many amino acids in a chain there would always be only one N-terminus group.
8 0
4 years ago
g Five calcite, CaCO3 (MW 100.085 g/mol), samples of equal mass have a total mass of 12.3±0.1 g. What is the absolute uncertaint
diamong [38]

Answer:

The value  is   L  =  0.985 \pm 0.00801 \  g

Explanation:

From the question we are told that

  The  molar mass of CaCO_3 is  MW  =  100.085 \  g/mol

   The  total mass is  m_g  = 12.3 \ g

   The uncertainty of the total mass is \Delta g  = 0.1

Generally the molar weight of calcium is M_c  =  40 g/mol

 The percentage of calcium in calcite is mathematically represented as

          C =  \frac{40.07}{100.085} * 100

          C =  40.03 \%

Generally the mass of each sample is mathematically represented as

     m=  \frac{m_g}{5}

     m=  \frac{12.3}{5}

     m= 2.46 \  g

Generally mass of calcium present in a single sample is mathematically represented as

        m_c = 2.46 *  \frac{40.04}{100}

       m_c = 0.985 \  g

The  uncertainty of  mass of a single sample is mathematically represented as

      k  =  \frac{\Delta g }{5}

        k  =  \frac{0.1 }{5}

       k  =  0.02\  g

The  uncertainty of  mass of calcium in a single sample is mathematically represent

         G  =  \frac{0.02 *  40.04}{ 100}

          G  =  0.00801 \  g

Generally the average mass of calcium in each sample is  

          L  =  0.985 \pm 0.00801

6 0
3 years ago
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