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3 years ago

Please help me thanks so much?!?.:;)...I’ll mark you.:;)))

Chemistry

2 answers:

faltersainse [42]3 years ago

8 0

Answer:

physical

Explanation:

oee [108]3 years ago

5 0

A physical change. When a physical change occurs the chemical composition of the substance doesn't change. Some examples of physical changes are melting, freezing, vaporization, and condensation.

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What is the number of valence electrons in a nitrogen atom in the ground state

ddd [48]

Answer: 5

Explanation:

It just is

6 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

What is the metal that is grouped with metalloids

vlabodo [156]

Answer:

The post-transition metals cluster to the lower left of this line. Metalloids: The metalloids are boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te) and polonium (Po). They form the staircase that represents the gradual transition from metals to nonmetals.

Explanation: i googled it

4 0

3 years ago

Use the given relationship to calculate the ionization energy of a thallium atom, given that radiation of 58.4 nm produces elect

Elenna [48]

The ionization energy is 1 × 10^-18 J

<h3>What is ionization energy?</h3>

The ionization energy is the energy required to remove an electron from an atom.

We have the following information;

wavelength of the photon = 58.4 nm

Speed of the electron = 2310 × 10^3 m/s

Since;

hv = I + 1/2mv^2

v = c/λ

hc/λ = I + 1/2mv^2

I = hc/λ - 1/2mv^2

I = (6.6 × 10^-34 × 3 × 10^8/58.4 × 10^-9) - (1/2 × 9.11 × 10^-31 × (2310 × 10^3)^2)

I = 1 × 10^-18 J

Learn more about ionization energy: brainly.com/question/16243729

3 0

2 years ago

A fact of nature that has been confirmed many times by observation is known as a(n).

BlackZzzverrR [31]

Answer:

law(scientific law)

Explanation:

8 0

4 years ago