What layer in the atmosphere has rain clouds

In which of these would a nonpolar covalent bond be present? HBr H2O HCI Br2

cluponka [151]

Br2 is the correct answer

5 0

3 years ago

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

KengaRu [80]

Answer:

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

5 0

3 years ago

2.088 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 4.746 g

uysha [10]

Answer:

The empirical formula is C3H6O

Explanation:

Step 1: Data given

Mass of the sample =2.088 grams

The mass contains carbon, hydrogen, and oxygen

Mass of CO2 produced = 4.746 grams

Mass of H2O produced = 1.943 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Atomic mass of C = 12.01 g/mol

Atomic mass of H = 1.01 g/mol

Atomic mass of O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 4.746 grams/ 44.01 g/mol

Moles CO2 = 0.1078 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.1078 moles CO2 we'll have 0.1078 moles C

Step 4: Calculate mass C

Mass C: moles C * atomic mass C

Mass C: 0.1078 moles * 12.01 g/mol

Mass C= 1.295 grams

Step 5: Calculate moles H2O

Moles H2O = 1.943 grams / 18.02 g/mol

Moles H2O = 0.1078 moles

Step 6: Calculate moles H

For 1 mol H2O we'll have 2 moles H

For 0.1023 moles H2O we'll have 2*0.1078 = 0.2156 moles H

Step 7: Calculate mass H

Mass H = 0.2046 moles * 1.01 g/mol

Mass H = 0.218 grams

Step 8: Calculate mass O

Mass O = 2.088 grams - 1.295 grams - 0.218 grams

Mass O = 0.575 grams

Step 9: Calculate moles O

Moles O = 0.575 grams / 16.0 g/mol

Moles O = 0.0359 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.1078 moles / 0.0359 moles = 3

H: 0.2156 moles / 0.0359 moles = 6

O: 0.0359 moles / 0.0359 moles =1

The empirical formula is C3H6O

8 0

3 years ago

This graph shows a relationship between the height of a giraffe and its weight. Which term describe the dashed line shown in the

Evgesh-ka [11]

Answer:

line of best fit I guss soo

8 0

3 years ago

Read 2 more answers

Heat will flow from a hot object to a cold object until the objects reach A. stability. B. boiling point. C. collapse. D. equili

Aleonysh [2.5K]

The thermal energy will flow from the hotter object to the cooler object until their temperatures are equal i.e. until they reach equilibrium.

Answer:
D. equilibrium

6 0

4 years ago

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