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The most common isotope of nitrogen is nitrogen-14, or 14N. Which of the following atoms has the same number of neutrons as 14N?

AlexFokin [52]

Here we have to choose the atom which has same number of neutrons as 14N

The atom which has same number of neutrons as 14N among the given atoms is (D) 13C

We know, in an atom the number of electrons (atomic number) is equivalent to the number of protons. For nitrogen (N), the number of electron is 7. Thus the number of protons present in the nucleus is also 7 to make the atom uncharged.

The mass number of the atom is the number of the neutrons + number of the protons. In 14N there are 7 protons and 7 neutrons.

Thus among the given atoms the atom which has 7 neutrons will have the same number of neutrons.

A) Now in 16O the atomic number is 8 and mass number is 16. Thus the number of protons and neutrons are each 8 units.

B) In 15N, the atomic number is 7 but mass number is 15, thus number of protons and neutrons are 7 and 8 respectively.

C) In 14P the atomic number is 15 but mass number is 14, so number of protons and neutrons are 15 and -1 (not exist) respectively.

D) In 13C the atomic number is 6 and mass number is 13. So the number of protons are 6 and number of neutrons are 7.

Thus the correct answer will be D) 13C.

3 0

4 years ago

Pls help me I beg you

just olya [345]

Answer:

Es la c

Explanation:

Espero que te ayude

6 0

3 years ago

Read 2 more answers

2 years ago a 4.00 l flexible container holds a sample of hydrogen gas at 153 kpa. if the pressure increases to 203 kpa and the

garri49 [273]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 x V1 / P2

V2 = 3 L

7 0

4 years ago

How much of a 0.250 M lithium hydroxide is required to neutralize 20.0 mL of 0.345M chlorous acid?

Bumek [7]

Answer:

27.6mL of LiOH 0.250M

Explanation:

The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:

LiOH + HClO₂ → LiClO₂ + H₂O

<em>That means, 1 mole of hydroxide reacts per mole of acid</em>

Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:

0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>

To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:

6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =

6 0

3 years ago

alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decima

svetlana [45]

A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$ = 0.10 M

The chemical equation for this reaction is :

$\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}$

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = $\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}$ = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

$\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}$

Initial 0.10 0.035 0.10 -

Change -0.035 -0.035 + 0.035 -

Equilibrium 0.065 0 0.135 -

By using Henderson-Hasselbalch equation:

The pH of this solution = pKa + log $\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}$

The pH of this solution = 4.74 + log $\mathtt{\dfrac{0.135}{0.065}}$

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06 to two decimal places

7 0

3 years ago