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3 years ago

The chemical reactivity increases for the

Chemistry

1 answer:

jekas [21]3 years ago

7 0

Answer:

D) The increase in size for the non-C atom

Explanation:

Among the Halogens, chemical reactivity decreases from F to I and also Atomic sizes increases from F to I. The bond length formed between C and The halogens decreases in strength from CF to CI which means the stronger the bond, the less reactive the compound.

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a gas occupies 600.0 ml at 20.00 degrees C and 70.00kPa what will be the pressure at 40.00 degrees C and 150.0mL?

Zina [86]

Answer:

P₂ = 299.11 KPa

Explanation:

Given data:

Initial volume = 600 mL

Initial pressure = 70.00 KPa

Initial temperature = 20 °C (20 +273 = 293 K)

Final temperature = 40°C (40+273 = 313 K)

Final volume = 150.0 mL

Final pressure = ?

Formula:

P₁V₁/T₁ = P₂V₂/T₂

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂

P₂ = 70 KPa × 600 mL × 313 K / 293K ×150 mL

P₂ = 13146000 KPa .mL. K /43950 K.mL

P₂ = 299.11 KPa

4 0

3 years ago

Does a reaction occur when aqueous solutions of copper(II) sulfate and lead(II) nitrate are combined?

Dafna11 [192]

In my opinion yes but thats me

6 0

3 years ago

A 21.82 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.33 grams of CO2 and 4

morpeh [17]

Answer: The molecular formula for the given organic compound is $C_2H_2O_4$

Explanation:

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 $CO_2=21.33g$

Mass of $H_2O=4.366g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 21.33 g of carbon dioxide, $\frac{12}{44}\times 21.33=5.82g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 4.366 g of water, $\frac{2}{18}\times 4.366=0.485g$ of hydrogen will be contained.

Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles$

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 0.485 moles.

For Carbon = $\frac{0.485}{0.485}=1$

For Hydrogen = $\frac{0.485}{0.485}=1$

For Oxygen = $\frac{0.969}{0.485}=1.99\approx 2$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_1H_{1}O_2=CHO_2$

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 = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

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

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Thus, the molecular formula for the given organic compound is $C_2H_2O_4$

4 0

4 years ago

How many SO2 molecules are in 1.76 mol of SO2? How many sulfur atoms and oxygen atoms are there?

Ugo [173]

Answer:

1.76 * 6.02*10^23 = 1.05952*10^24

1.05952*2 = 2.11904 *10^24 oxygen and 1.05952*10^24 sulfur atoms

5 0

2 years ago

What is the net ionic equation of H3PO4(aq) + Ca(OH)2(aq) yields Ca3(PO4)2(aq) + H2O(l)

aalyn [17]

Answer:

The net ionic equation is:

6 OH⁻(aq) + 6 H⁺(aq) ⇒ 6 H₂O (l)

Explanation:

First it is convenient to determine the cation and the anion in each compound. Cations are the positive ions of a compound and are usually metals. Anions are the negative ions of a compound and are nonmetals.For that it is important to recognize polyatomic ions in a reaction. Polyatomic ions are charged molecules that bind so strongly that they cannot disassociate during chemical reactions. In other words, polyatomic ions are groups of two or more atoms that are chemically linked. These have a specific load and do not separate into individual components. Polyatomic ions can have both positive and negative charges. Some of the most common polyatomic ions are CO₃²⁻, NO₃⁻, NO₂⁻, SO₄²⁻, SO₃²⁻, ClO₃⁻, PO₄³⁻ and OH⁻.

Then you must make sure that the initial equation is completely balanced. To balance an equation, you must add coefficients in front of the compounds to match the amount of atoms of each element on both sides of the equation.

In this case, the final balanced equation after balancing is:

3 Ca(OH)₂(aq) + 2 H₃PO₄(aq) ⇒ 6 H₂O(l) + Ca₃(PO₄)₂(s)

When a species or compound dissociates, it separates into its positive (cation) and negative (anion) components. These will be the components that are balanced at the end for the net ionic equation.

Given that strong acids such as H₃PO₄ and strong bases such as Ca(OH)₂ are completely dissociated, in addition to everything previously mentioned related to polyatomic ions, the net ionic equation will be:

3 Ca⁺² (aq) + 6 OH⁻(aq) + 6 H⁺(aq) + 2 PO₄⁻³(aq) ⇒ 6 H₂O (l)+3 Ca⁺² (aq) + 2 PO₄⁻³ (aq)

In this case it was taken into account that the coefficients located in front of each molecule indicate the amount of each molecule for the reaction and that the subscripts next to each atom or ion is the number of atoms or ions in the equation. By multiplying the coefficient mentioned by the subscript, you get the amount of each element or ion present in the reaction.

The spectator ions are those that are present on both sides of the reaction and are not changed by the reaction, that is, they are present in exactly the same amount on both sides of the equation. In this way, they can be canceled from both sides of the reaction. The spectator ions in this case are PO₄³⁻(aq) and Ca²⁺(aq).

This leaves you with the net ionic equation:

6 OH⁻(aq) + 6 H⁺(aq) ⇒ 6 H₂O (l)

7 0

3 years ago