The simplest formula for a compound made from element X(molar mass=79.0g mol) that is 21.0% nitrogen by mass is __

Give the formula and name for the iconic compounds formed from the following pairs of elements:

Fudgin [204]

(1) Give the formula and name for the iconic compounds formed from the following pairs of elements:

A. KCl potassium chloride

B. ZnO zinc oxide

(2) Give the formula for each of the following ionic compounds:

A. KClO₃ potassium chlorate

B. MgSO₄ magnesium sulfate

C. KNO₃ potassium nitrate

(3) Give the name of the following covalent compounds:

A. SO₂ sulfur dioxide

B. CCl₄ carbon tetrachloride

C. SiS₂ silicon disulfide

D. SCl₂ sulfur dichloride

4 0

3 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

Answer: The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of $CO_2=3.926g$

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

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

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

Step 2: 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 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is $CH_2F$

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 = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

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

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago

Consider the chemical equation. CuCl2 + 2NaNO3 Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with

Finger [1]

Answer:

The percent yield of NaCl is 78.7 %

Explanation:

CuCl₂ + 2NaNO₃ → Cu(NO₃)₂ + 2NaCl

If the NaNO₃ is determined to be in excess, the limiting reagent is the chloride. We convert the mass to moles:

31 g . 1mol / 134.45g = 0.230 moles

Ratio is 1:2, so we can make a rule of three to determine the theoretical yield

1 mol of copper (II) chloride reacts to produce 2 moles of sodium chloride

Then, 0.230 moles of CuCl₂ will react to produce (0.230 .2) /1 ) = 0.461 moles of NaCl → we convert the moles to mass → 0.461 mol . 58.45 g / 1mol = 26.9 g

To find percent yield we do → (Yield produced / Theoretical yield) . 100

(21.2 g / 26.9 g) . 100 = 78.7 %

3 0

3 years ago

Will NaCl be soluble or insoluble

attashe74 [19]

NaCl is salt to it is obviously SOLUBLE :)

6 0

4 years ago

What happens to the particles of a solid when the solid changes into a liquid? (5 points) a They increase in size. b They decrea

tatiyna

The answer is D, the attractive force between them decreases.

Although the volume that the particles take up generally increases when going from solid to liquid, particles themselves never change their size. It’s the attractive force that decreases and doesn’t pull the particles together as tightly when a solid changes to liquid.

6 0

3 years ago

The simplest formula for a compound made from element X(molar mass=79.0g mol) that is 21.0% nitrogen by mass is ___