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3 years ago

What charge would an ion have if it had more electrons than protong?

Chemistry

1 answer:

max2010maxim [7]3 years ago

4 0

Answer:

D. It would have a net negative charge

Explanation:

In an atom, there are electrons, protons and neutrons collectively called SUBATOMIC PARTICLES. The charge of an atom is determined by the number of electrons and protons in that atom. A neutral or uncharged atom possess the same number of electrons and protons.

The charge can thus be calculated as follows:

Charge of atom = no. of protons - no. of electrons.

Hence, when the number of electrons in an atom is higher than that of the protons, it means that the atom has gained negatively charged electrons, making that atom have a net negative charge. For example, In an atom that contains 10 protons and 11 electrons, the charge of that atom will be (10 - 11) = -1. The net charge of that atom will be -1, which is NEGATIVE.

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

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

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

7 When carbon is heated in a limited supply

irakobra [83]

Answer:

please mark my answer brainliest

Explanation:

its carbon monoxide

.....

5 0

4 years ago

Read 2 more answers

Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol

zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = $10^{-4.8}$

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = $10^{-7.6}$

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = $10^{-1.9}$

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = $10^{1.8}$

[In-]/[HIn] = 63.10

6 0

3 years ago

Part A

Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

7.33 × 10 ⁻³ c

Explanation:

1) Data:

a) m = 9.11 × 10⁻³¹ kg

b) λ = 3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

2) Formula:

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

λ = h / (m.s)

h is Planck's constant: h= 6.626×10⁻³⁴J.s

3) Solution:

Solve for s:

s = h / (m.λ)

Substitute:

s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s

s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

2.06 × 10 ⁻³⁴ m.

Explanation:

1) Data:

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ = ?

2) Formula:

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

λ = h / (m.s)

h is Planck's constant: h= 6.626×10⁻³⁴J.s

3) Solution:

λ = h / (m.s)

Substitute:

λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.

5 0

3 years ago

A solution has a [OH-] of 1 × 10-9. What is the pOH of this solution?

zhannawk [14.2K]

POH = - log [ OH⁻ ]

pOH = - log [ 1 x 10⁻⁹ ]

pOH = 9

Answer C

hope this helps!

3 0

3 years ago