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slega [8]
3 years ago
12

Which ion has the largest radius?1) Br-2) Cl-3) F-4) I-​

Chemistry
1 answer:
Zanzabum3 years ago
3 0
I think the answer is 1
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An iron chloride compound contains 55.85 grams of iron and 106.5 grams of chlorine. What is the most likely empirical formula fo
mart [117]

Answer:

FeCl_{3}

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

mass of Fe = 55.85 g

Molar mass of Fe = 55.85 g/mol

<u>Moles of Fe = 55.85 / 55.85 = 1</u>

mass of Cl = 106.5 g

Molar mass of Cl = 35.5 g/mol

Moles of Cl = 106.5 / 35.5 = 3

Taking the simplest ratio for Fe and Cl as:

1 : 3

The empirical formula is = FeCl_{3}

3 0
3 years ago
What is one reason why rock layers are not horizontal?
lawyer [7]
B)a force pushed the rock layers after they were formed
6 0
3 years ago
Read 2 more answers
The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its
Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

t_{\frac{1}{2}} =\frac{ln 2}{k}

t_{\frac{1}{2}} = \frac{0.693}{k}

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}

k = \frac{0.693}{5230 years}

k = 1.325 * 10^{-4} yr^{-1}

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

[A] = [A_{0}]e^{-kt}

\frac{[A]}{[A_{0}]} = e^{-kt}

\frac{35}{100} = e^{-(1.325 *10^{-4})t}

ln(0.35) = -(1.325 *10^{-4})(t)

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0
3 years ago
When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

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

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

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

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <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 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

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

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

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 = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

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

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

6 0
3 years ago
There are two opposing processes that occur in a solution in contact with undissolved solute. These are dissolving and ________.
Alex777 [14]

Answer:

c. crystallization

Explanation:

The opposing process that occur in a solution in contact with undissolved solute are dissolution and crystallization.

In the dissolution process the solid substance coverts into liquid state and mixes with solution. Whereas in Crystallization the the chemical is converted from the liquid solution to solid crystal state.

3 0
3 years ago
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