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lianna [129]
2 years ago
15

Calculate δh o for the reaction. enter your answer in the provided box. ch3oh + hf → ch3f + h2o

Chemistry
1 answer:
Vaselesa [24]2 years ago
3 0

Answer:- delta H for the given reaction is -12.6 kJ/mol.

Solution:- delta H of the reaction = sum of standard enthalpy of formation of products - sum of standard enthalpy of formation of reactants

Standard enthalpy of formation values could be found in thermodynamic data table given in book.

The values for methanol, hydrofluoric acid, methylfluoride and water are -238.6 kJ/mol, -268.6 kJ/mol, -234kJ/mol and -285.8 kJ/mol respectively.

Please note, the phases taken for methanol, hydrofluoric acid, methylfluoride and water are liquid, gas, gas and liquid respectively.

Let's plug in the values in the formula.

delta H for the reaction = [(-234) +(-285.8)] - [(-238.6) + (-268.6)]

delta H for the reaction = -519.8 + 507.2

delta H for the reaction = -12.6 kJ/mol

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Which of the following statements about monosaccharide structure is true?
Nana76 [90]

Answer:

The only statement about monosaccharide structure which is true is b. (Monosaccharides can be classified according to the spatial arrangement of their atoms)

Explanation:

Monosaccharides are simple sugars that are classified according to the amount of carbon atoms and based on these numbers, we can call them trioses, pentoses and hexoses. They are molecules with aldehyde (aldose) or centone (ketose) groups that have more than one alcohol function, but which do not differ in their position (OH). They do not contain N, since their general formula is Cx (H2O) x. A 6-carbon monosaccharide is called hexose, since the pentose only has 5

8 0
2 years ago
Write the electron configurations for the following ions:
Ket [755]

Answer:

Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Sn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zr⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Ag⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

S²⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶

Explanation:

Cobalt (Co): atomic number 27

<u>The electronic configuration of Co in ground state: </u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷

<u>The electronic configuration of Co in +2 oxidation state (Co²⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Tin (Sn): atomic number 50

<u>The electronic configuration of Sn in ground state: </u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²

<u>The electronic configuration of Sn in +2 oxidation state (Sn²⁺) </u>:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zirconium (Zr): atomic number 40

<u>The electronic configuration of Zr in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²

<u>The electronic configuration of Zr in +4 oxidation state (Zr⁴⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Silver (Ag): atomic number 47

<u>The electronic configuration of Ag in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰

<u>The electronic configuration of Ag in +1 oxidation state (Ag⁺) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

Sulphur (S): atomic number 16

<u>The electronic configuration of S in ground state:</u>

1s² 2s² 2p⁶ 3s² 3p⁴

<u>The electronic configuration of S in -2 oxidation state (S²⁻) :</u>

1s² 2s² 2p⁶ 3s² 3p⁶

8 0
2 years ago
Read 2 more answers
A. A nurse practitioner prepares 500. mL of an IV of normal saline solution to be delivered at a rate of 80. mL/h. What is the i
nydimaria [60]
A. Quantity of saline = 500mL 
Rate of infusion = 80 mL / h 
Infusion time = Quantity / Rate = 500 mL / (80 mL/hr) = 6.25 hr 
b. Child weight = 72.6 lb = 32.93 kg 
Medrol to be given = 1.5 mg per kg 
Quantity of Medrol = 20 mg/mL 
Dosage available = 20 mg/mL / 1.5 mg/kg = 13.33 kg/mL 
Dosage according to body weight = 32.93 kg / 13.33 kg/mL = 2.47 mL
8 0
2 years ago
A 126-gram sample of titanium metal is heated from 20.0°C to 45.4°C while absorbing 1.68 kJ of heat. What is the specific heat o
Radda [10]

Answer:

The specific heat for the titanium metal is 0.524 J/g°C.

Explanation:

Given,

Q = 1.68 kJ   = 1680 Joules

mass = 126 grams

T₁ = 20°C

T₂ = 45.4°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

Here, ΔT =  T₂ - T₁ = 45.4 - 20 = 25.4°C.

Substituting values,

1680 = (126)(25.4)(Cp)

By solving,

Cp = 0.524 J/g°C.

The specific heat for the titanium metal is 0.524 J/g°C.

3 0
2 years ago
A chemical engineer calculated that 15.0 mol H2 was needed to react with excess N2 to prepare 10.0 mol NH3. But the actual yield
rjkz [21]

Answer:

The actual number of moles is 9 moles.

It is less than 15

Number of moles needed is 9 moles

Explanation:

15H2 + 10N2 ——-> 10NH3

Now from the question, we can see that the percentage yield is 60%

The percentage yield can be calculated as actual moles of H2/Theoretical moles of H2 * 100%

From the equation, we can see that the theoretical number of moles of hydrogen is 15.

Now to get the actual : 60 = x/15 * 100

x = 9

The actual number of moles is 9 moles.

It is less than 15

Number of moles needed is 9 moles

8 0
2 years ago
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