Ask question

Ask question

All categories

3 years ago

5

In the lab vinegar acetic acid was mixed with baking soda sodium bicarbonate to form dioxide gas and sodium acetate. What are th

e products?

1 answer:

SpyIntel [72]3 years ago

4 0

2CH3COOH +Na2CO3 ----> 2CH3COONa + H20 + CO2

know you can find what all products formed from his reaction

You might be interested in

If an old car burns 0.05 mL oil every mile, How much oil will it burn if driven 100,000 miles? Show your answer in mL and L?.

Irina18 [472]

0.05 * 100000

= 5000ml

= 5L

7 0

3 years ago

How are atoms connected to Electricity

qwelly [4]

The protons and electrons of an atom are attracted to each other. They both carry an electrical charge. Protons have a positive charge (+) and electrons have a negative charge (-). The positive charge of the protons is equal to the negative charge of the electrons.

5 0

2 years ago

Fe(s) + CuSO4(aq) ⇒ Cu(s) + FeSO4(aq)

Alexxandr [17]

Answer:

The answer to your question is CuSO₄

Explanation:

To answer your question just remember the following information

- A chemical reaction is divided into sections

reactants and products

reactants on the left side of the reaction

products on the right side of the reaction

- All the symbols have a meaning

(s) means that that compound is in solid phase

(aq) means that that compound is dissolved in solution.

Then, the answer is CuSO₄

4 0

3 years ago

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

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

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

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

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

3 years ago

Olivia bought new gym shoes to play volleyball because she kept slipping when she ran in her old shoes. How will the new soles h

HACTEHA [7]

Maybe her old shoes had soft worn out bottoms and she slips in them. So her new shoes had more grip than her old ones so they kept her from falling.

6 0

2 years ago

Read 2 more answers