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2 years ago

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Match the element with its number of valence electrons. Match Term Definition: Helium Aluminum Fluorine Sodium A) 1 B) 2 C) 3 D)

7

2 answers:

wariber [46]2 years ago

5 0

Helium- 2

Aluminum- 3

Fluorine- 7

Sodium- 1

Hope I helped.

vampirchik [111]2 years ago

3 0

<u>Explanation:</u>

Valence electrons are defined as the electrons that are present in the outermost shell of an atom.

Helium is the 2nd element of the periodic table having electronic configuration of $1s^2$

This element has 2 electrons in its outermost shell. So, it has 2 valence electrons. The correct answer is Option B.

Aluminium is the 13th element of the periodic table having electronic configuration of $1s^22s^22p^63s^23p^1$

This element has 3 electrons in its outermost shell. So, it has 3 valence electrons. The correct answer is Option C.

Fluorine is the 9th element of the periodic table having electronic configuration of $1s^22s^22p^5$

This element has 7 electrons in its outermost shell. So, it has 7 valence electrons. The correct answer is Option D.

Sodium is the 11th element of the periodic table having electronic configuration of $1s^22s^22p^63s^1$

This element has 1 electron in its outermost shell. So, it has 1 valence electron. The correct answer is Option A.

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

3 years ago

7.02 x 10^23 molecules of X2, a ssubstance consisting of diatomic molecules, has a mass of 296 grams. Determine the atomic weigh

Fiesta28 [93]

Answer:

d. 127 g/mol.

Explanation:

Hello!

In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:

$mol=7.02x10^{23}molec*\frac{1mol}{6.022x10^{23}molec}=1.17mol$

Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:

$MM=\frac{296g}{1.17mol}\\\\MM=254g/mol$

It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.

Best regards!

4 0

2 years ago

Definition: This is a net gain or loss of electrons.

AlekseyPX

Answer:

Elecric charge/ Electricity

Explanation:

Electric charge is the net gain or loss of electrons

I hope im right!!

3 0

2 years ago

Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg C. The densities of hexane and octane are 0.655 g/cm3 an

avanturin [10]

Answer:

The required volume of hexane is 0.66245 Liters.

Explanation:

Volume of octane = v=1.0 L= $1000 cm^3$

Density of octane= d = $0.703 g/cm^3$

Mass of octane ,m= $d\times v=0.703 g/cm^3\times 1000 cm^3=703 g$

Moles of octane = $\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol$

Mole percentage of Hexane = 45%

Mole percentage of octane = 100% - 45% = 55%

$55\%=\frac{6.166 mol}{\text{Total moles}}\times 100$

Total moles = 11.212 mol

Moles of hexane :

$45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100$

Moles of hexane = 5.0454 mol

Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g

Density of the hexane,D = $0.655 g/cm^3$

Volume of hexane = V

$V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L$

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.

5 0

3 years ago