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2 years ago

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Elements in group 1 will (lose or gain) electrons to obtain a noble gas structure. How many electrons will the element gain or l

ose in order to obtain a noble gas structure? (5,3,7,6,1,2,4)

1 answer:

Whitepunk [10]2 years ago

3 0

Answer is: elements in group 1 will lose electrons to obtain a noble gas structure. They will lose 1 electron.
For example ₃Li 1s²2s¹ will lose one electron from 2s oribtal to obtain helium structure ₂He 1s².
Or sodium ₁₁Na 1s²2s²2p⁶3s¹ will also lose one electron to obtain neon structure ₁₀Ne 1s²2s²2p⁶.

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One isotope of oxygen differs from another isotope of oxygen in *

Reptile [31]

Each isotope of Oxygen has a different number of neutrons

<h3>Further explanation </h3>

The elements in nature have several types of isotopes

Atomic mass is the average atomic mass of all its isotopes

Isotopes are atoms has the same number of protons but has a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Some of the isotopes of oxygen are:

$\tt _8^{16}O,_8^{17}O,_8^{18}O$

Each isotope has 8 protons and 8 electrons but has a different number of neutrons

For O-16: number of neutrons = 16-8 = 8

For O-17: number of neutrons = 17-8 = 9

For O-18: number of neutrons = 18-8 = 10

4 0

2 years ago

Question 13 of 25

avanturin [10]

Answer: is A

Explanation:

how becuase i just take the test

7 0

2 years ago

If you have 23.8g of CaCl2, how many formula units is it?

NemiM [27]

Answer:

1.3×10²³ formula unit

Explanation:

Given data:

Mass of CaCl₂ = 23.8 g

Number of formula unit = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 23.8 g/110.98 g/mol

Number of moles = 0.21 mol

1 mole of any substance contain 6.022×10²³ formula unit

0.21 mol × 6.022×10²³ formula unit / 1mol

1.3×10²³ formula unit

8 0

2 years ago

Determine the empirical formulas for compounds with the following percent compositions:

Elis [28]

<u>Answer:</u>

<u>For a:</u> The empirical formula of the compound is $P_2O_5$

<u>For b:</u> The empirical formula of the compound is $KH_2PO_4$

<u>Explanation:</u>

<u>For a:</u>

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.

For Phosphorus = $\frac{1.406}{1.406}=1$

For Oxygen = $\frac{3.525}{1.406}=2.5$

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = $1\times 2=2$

For Oxygen = $2.5\times 2=5$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is $P_2O_5$

<u>For b:</u>

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles$

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.

For Potassium = $\frac{0.736}{0.735}=1$

For Hydrogen = $\frac{1.5}{0.735}=2.04\approx 2$

For Phosphorus = $\frac{0.735}{0.735}=1$

For Oxygen = $\frac{2.9375}{0.735}=3.99\approx 4$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is $KH_2PO_4$

3 0

3 years ago

One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas?

Andre45 [30]

Explanation:

The given data is as follows.

n = 1 mol, $V_{f} = 2V_{i}$

Q = 1500 J, R = 8.314 J/mol k

(a) $\Delta S = \frac{dQ}{dT}$

And, according to the first law of thermodynamics

$\Delta E_{int} = Q - W$

And, in an isothermal process the change in internal energy of the gas is zero.

Hence, 0 = Q - W

or, W = Q

Expression for work done in an isothermal process is as follows.

W = $nRT ln \frac{V_{f}}{V_{i}}$

As W = Q, Hence expression for Q will also be given as follows.

Q = $nRT ln \frac{V_{f}}{V_{i}}$

Now,

$\Delta S = \frac{nRT ln \frac{V_{f}}{V_{i}}}{T}$

[/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]

= $nR ln \frac{2V_{i}}{V_{i}}$

= nR ln 2

= $1 \times 8.314 \times 0.693$

= 5.76 J/K

Therefore, change in entropy is 5.76 J/K.

(b) As, Q = $nRT ln \frac{V_{f}}{V_{i}}$

= $nRT ln \frac{2V_{i}}{V_{i}}$

= nRT ln 2

T = $\frac{Q}{nR ln 2}$

= $\frac{1500}{1 \times 8.314 ln 2}$

= 260.4 K

Therefore, temperature of the gas is 260.4 K.

7 0

2 years ago