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Eddi Din [679]
3 years ago
10

You will analyze four substances in this lab. Based on their chemical formulas and what you already know about covalent and ioni

c compounds, make a prediction for each compound.
Oil, which is built from the nonmetals hydrogen, carbon, and oxygen:
Cornstarch, a carbohydrate consisting of hydrogen, carbon, and oxygen:
Sodium chloride (table salt), whose formula is NaCl:
Sodium bicarbonate (sodium bicarbonate), whose formula is NaHCO3:
Chemistry
2 answers:
Phoenix [80]3 years ago
7 0

Answer: Oil: covalent

Cornstarch: Covalent

Sodium chloride: Ionic

Sodium bicarbonate: Ionic

Explanation: Covalent compounds are formed by sharing of electrons between non metals whereas ionic compounds are formed by transfer of electrons from metals to non metals.

1. Oil, which is built from the nonmetals hydrogen, carbon, and oxygen: forms a covalent compound by sharing of electrons between non metals hydrogen, carbon, and oxygen. Covalent compounds are insoluble in water.

2. Cornstarch, a carbohydrate consisting of hydrogen, carbon, and oxygen: forms a covalent compound by sharing of electrons between non metals hydrogen, carbon, and oxygen. Covalent compounds are insoluble in water.

3. Sodium chloride (table salt), whose formula is NaCl is formed by transfer of electrons from sodium to chlorine.Ionic compounds are soluble in water.

4. Sodium bicarbonate, whose formula is NaHCO_3 is formed by transfer of electrons from sodium to HCO_3.Ionic compounds are soluble in water.

marin [14]3 years ago
3 0

Answer:

Answer: Oil: covalent

Cornstarch: Covalent

Sodium chloride: Ionic

Sodium bicarbonate: Ionic

Explanation:

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valina [46]

Answer: Option (b) is the correct answer.

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3 years ago
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The rate constants of some reactions double with every 10-degree rise in temperature. Assume that a reaction takes place at 295
LUCKY_DIMON [66]

Answer : The activation energy for the reaction is, 51.9 kJ

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 295 K

K_2 = rate constant at 305 K = 2K_1

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 295 K

T_2 = final temperature = 305 K

Now put all the given values in this formula, we get:

\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{295K}-\frac{1}{305K}]

Ea=51879.96J=51.9kJ

Therefore, the activation energy for the reaction is, 51.9 kJ

8 0
4 years ago
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