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2 years ago

Give the following number of protons, neutrons, and electrons in the following isotope:

Chemistry

1 answer:

valina [46]2 years ago

3 0

Answer:

47, 68, 47

Explanation:

Let's consider the following isotope: ¹¹⁵Ag.

If we look in the Periodic Table, the atomic number of Ag is 47, that is, silver has 47 protons.

Since ¹¹⁵Ag is neutral it must have the same number of protons that of electrons. Then, it also has 47 electrons.

The number 115 in ¹¹⁵Ag stands for the mass number, which is the sum of protons and neutrons. Then, the number of neutrons is:

p⁺ + n⁰ = 115

n⁰ = 115 - p⁺ = 115 - 47 = 68

The number of protons, neutrons, and electrons in ¹¹⁵Ag is: 47, 68, 47.

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Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.

Misha Larkins [42]

Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

$moles HCl = moles KOH \\ \\ cHCl*vHCl=cKOH*vKOH \\ \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}= 541,54mL$

So, the required volume of HCl is 541,54 mL

Have a nice day!

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Goryan [66]

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2 years ago

Draw the structure of a compound with the molecular formula C9H10O2 that exhibits the following spectral data. (a) IR: 3005 cm-1

Nikitich [7]

Answer:

The answer is 3-Phenylpropanoic acid (see attached structure)

Explanation:

From spectral data:

3005 cm-1 ⇒ carboxylic acid (broad band)

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1603 cm-1 ⇒ Aromatic C-C bond

H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding (clearer investigation of spectrum would be expedient).

Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.

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Explanation:

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Top period will have the atom with larger radius.

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2 years ago