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Blizzard [7]
3 years ago
13

Classify the chemical equations as being balanced or not balanced. 2CO 2NO → 2CO2 N2 6CO2 6H2O → C6H12O6 O2 H2CO3 → H2O CO2 2Cu

O2 → CuO.
Chemistry
1 answer:
marissa [1.9K]3 years ago
4 0

Answer:

2CO 2NO → 2CO2 N2 : Balanced

6CO2 6H2O → C6H12O6 : Unbalanced

H2CO3 → H2O CO2 : Balanced

2Cu O2 → CuO : Unbalanced

Explanation:

1.) 2CO 2NO → 2CO2 N2

2 Carbon 2

4 Oxygen 4

2 Nitrogen 2

The amount of atoms of each element on each side of the equation are the same therefore the equation is balanced.

2.) 6CO2 6H2O → C6H12O6 O2

6 Carbon 6

12 Oxygen 8

12 Hydrogen 12

The amount of oxygen atoms is different on both sides of the equation therefore the equation is not balanced.

3.) H2CO3 → H2O CO2

2 Hydrogen 2

1 Carbon 1

3 Oxygen 3

The amount of atoms of each element is the same on both sides of the equation therefore the equation is balanced.

2Cu O2 → CuO.

2 Cu 2

2 O 1

The amount of oxygen atoms is different on both sides of the equation therefore the equation is not balanced.

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2. An exothermic reaction releases 86.5 kJ. How many kilocalories of<br> energy are released?
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20.67 kcal of energy is released.

Explanation:

It is given that, an exothermic reaction releases 86.5 kJ. We need to convet kJ to calories.  

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3 years ago
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4 years ago
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Calculate the vapor pressure of spherical water droplets of radius (a) 17 nm and (b) 2.0 μm surrounded by water vapor at 298 K.
Montano1993 [528]

Explanation:

Relation between pressure of water and its droplet is as follows.

           ln (\frac{p}{p_{o}}) = \frac{2 \gamma M}{r \rho RT}

where,   p = pressure of droplet

          p_{o} = water pressure in given temperature

          \gamma = 7.99 \times 10^{-3}

           M = Molecular Weight in Kg/Mol (0.018 for water)

            r = radius in meters

     \rho = density of water in Kg/m^{3} (1000 kg/m^{3})

           R = ideal gas constant (8.31)

           T = temperature in Kelvin

(a)   We will calculate the value of p as follows.

           p = e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}

              = e^{\frac{2 \times 0.07199 \times 0.018}{1.7 \times 10^{-8} \times 1000 \times 8.31 \times 298 K} \times 25.2

              = 26.8 torr

(b)  And, vapor pressure of spherical water droplets of radius 2.0 \mu m or 2 \times 10^{-6} m

             p = e^{\frac{2 \gamma M}{r \rho RT}} \times p_{o}

              = e^{\frac{2 \times 0.07199 \times 0.018}{2 \times 10^{-6} \times 1000 \times 8.31 \times 298 K} \times 25.2

              = 25.2 torr

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