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satela [25.4K]
2 years ago
12

What is/are the product(s) of a neutralization reaction of a carboxylic acid? View Available Hint(s) What is/are the product(s)

of a neutralization reaction of a carboxylic acid? a neutral compound and water a carboxylate salt and water a carboxylate salt a base and water g
Chemistry
1 answer:
lutik1710 [3]2 years ago
7 0

Answer:

A carboxylate salt and water

Explanation:

A carboxylic acid is an organic compound that has general formula RCOOH, where R is a carbon chain. Because it's an acid, the neutralization will happen when it reacts with a base, such as NaOH.

When this reaction occurs, the base will dissociate in Na⁺ and OH⁻, and the acid will ionize in RCOO⁻ and H⁺, so the products will be RCOO⁻Na⁺ (a carboxylate salt) and H₂O (water).

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EleoNora [17]

Answer:

54 g

Explanation:

1 mole of water = H2O

mass of 1 mole of H2O= mass of h2 + mass of o

= 2× mass of h +mass of o

= 2×1+16 =18 g

1 mole of water = 18g

3moles of water = 18×3g= 54g

6 0
3 years ago
What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme
boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 1.096776\times 10^7m^{-1}

n_f = Higher energy level = n_i+1=(2+1)=3

n_i= Lower energy level = 2    (Balmer series)

Putting the values in above equation, we get:

\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m

Converting this into nanometers, we use the conversion factor:

1m=10^9nm

So, 6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm

Hence, the longest wavelength of light is 656.5 nm

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2 years ago
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Answer:

They are both listed under group 11 on the periodic table and both are highly conductive of electricity

Explanation:

HOPE THIS HELPS ^^

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The correct answer is B.LONGITUDINAL WAVES.
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