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1. Take a look at all the elements in period 3. What do you notice? *

Vedmedyk [2.9K]

Answer:

D

Explanation:

beacause all the elements in period 3 has different valence electrons hence you look at the first element in period 3 is sodium which has a valence electron of 1 when you go across the period to the right their valence electron decreases to be negative

5 0

2 years ago

Read 2 more answers

8). How many moles of oxygen gas must be placed in a 35.0 L container in order

klemol [59]

Answer:

4.1 moles

Explanation:

Applying

PV = nRT................ equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = Temperature.

make n the subject of the equation

n = PV/RT.............. Equation 2

From the question,

Given: V = 35 L , P = 2.8 atm, T = 15 °C = (15+273) = 288 K, R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (35×2.8)/(0.083×288)

n = 4.1 moles

5 0

2 years ago

Which of the following SI or metric units is a base unit?

Olin [163]

3. Ampere is the base unit for electric current.

8 0

3 years ago

Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03%

nlexa [21]

<u>Answer:</u> The empirical and molecular formula of chrysotile is $Mg_3Si_2H_3O_4$ and $Mg_6Si_4H_6O_{16}$

<u>Explanation:</u>

We are given:

Percentage of Mg = 28.03 %

Percentage of Si = 21.60 %

Percentage of H = 1.16 %

Percentage of O = 49.21 %

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

Mass of Mg = 28.03 g

Mass of Si = 21.60 g

Mass of H = 1.16 g

Mass of O = 49.21 g

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

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

Moles of Magnesium = $\frac{\text{Given mass of Magnesium}}{\text{Molar mass of Magnesium}}=\frac{28.03g}{24g/mole}=1.17moles$

Moles of Silicon = $\frac{\text{Given mass of Silicon}}{\text{Molar mass of Silicon}}=\frac{21.06g}{28g/mole}=0.752moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.16g}{1g/mole}=1.16moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{49.21g}{16g/mole}=3.07moles$

<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 0.752 moles.

For Magnesium = $\frac{1.17}{0.752}=1.5$

For Silicon = $\frac{0.752}{0.752}=1$

For Hydrogen = $\frac{1.16}{0.752}=1.5$

For Oxygen = $\frac{3.07}{0.485}=4.08\approx 4$

To convert the mole ratios into whole numbers, we multiply individual mole ratios by 2

Mole ratio of Magnesium = (2 × 1.5) = 3

Mole ratio of Silicon = (2 × 1) = 2

Mole ratio of Hydrogen = (2 × 1.5) = 3

Mole ratio of Oxygen = (2 × 4) = 8

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

The ratio of Mg : Si : H : O = 3 : 2 : 3 : 8

The empirical formula for the given compound is $Mg_3Si_2H_3O_8$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 520.8 g/mol

Mass of empirical formula = [(24 × 3) + (28 × 2) + (1 × 3) + (16 × 8)] = 259 g/mol

Putting values in above equation, we get:

$n=\frac{520.8g/mol}{259g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$Mg_{(3\times 2)}Si_{(2\times 2)}H_{(3\times 2)}O_{(8\times 2)}=Mg_6Si_4H_6O_{16}$

Hence, the empirical and molecular formula of chrysotile is $Mg_3Si_2H_3O_4$ and $Mg_6Si_4H_6O_{16}$

5 0

3 years ago

Calculate the partial pressure of oxygen given the total barometric pressure of 515 mmHg. Round to the nearest hundredth.

Eva8 [605]

The partial pressure of oxygen given the total barometric pressure is : 108.15 mmHg

<u>Given data : </u>

Total barometric pressure = 515 mmHg

Assuming oxygen percentage = 21%

Barometric pressure dry at 37°C

<h3 /><h3>Determine the partial pressure of oxygen </h3>

Applying the relation below

Partial pressure = oxygen percentage * Barometric pressure

= 21% * 515 mmHg

= 108.15 mmHg

Hence we can conclude that the partial pressure of oxygen is 108.15 mmHg.

Learn more about Partial pressure : brainly.com/question/1835226

8 0

2 years ago