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3 years ago

In which of the following choices is the oxidation number incorrect?

Chemistry

1 answer:

8_murik_8 [283]3 years ago

4 0

Cl2(s); oxidation number 1 is the incorrect choices in oxidation number.

Explanation:

In the elemental form oxidation state is zero. Here chlorine is present in elemental form so oxidation state is zero.

Oxidation number depends on the number of electrons gained or lost by an atom of the element say in compound formation.

If electron is gained oxidation number becomes negative.

If electron is lost then oxidation number is positive.

If the octet rule is fulfilled that valence shell is filled them atomic number gets zero. Since Cl2 is in neutral state the oxidation number is 0.

Oxidation number in general can be made out by checking the valency of the element as oxidation number is also equal to the valency.

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Aluminum reacts with oxygen to produce aluminum oxide. If I give Tanya 3.2 grams of aluminum, how many g of oxygen does she need

timurjin [86]

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3 years ago

When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

4 years ago

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Alinara [238K]

Answer:

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Explanation:

3 0

3 years ago

Two more substances put together but each retaining their own identity is a

Lana71 [14]

The answer is Mixture. Two more substances put together but each retaining their own identity is a Mixture. They are two or more substances dispersed in one another but each retaining their own identity.

6 0

4 years ago

The heat equation, as given in the introduction, can also be rearranged to calculate the mass or temperature change for a substa

notka56 [123]

Answer:

The correct answer is 281.39 grams.

Explanation:

To arrive at this answer you must first keep in mind the basic equation:

Q = m*Cp* ΔT

Now, in order to calculate the necessary aluminum mass that absorbs 2138 J when passing from 14.83 to 23.31 ° C you must "clear" m of the previous equation.

This means, leave only the mass on one side of the equation, and "pass" Cp and ΔT to the other side dividing Q. This would look like this:

m= Q/ (Cp*ΔT)

Then, you need the value of specific heat of aluminum in the correct units, that is J / g ° C, the approximate value is 0.896.

ΔT is calculated by doing the mathematical operation:

23.31 °C - 14.83 °C = 8.48 °C

Finally, the values of: Q (data provided in joules), Cp (J / g ° C) and ΔT (calculated in ° C) are replaced in the last equation and the mass (in grams) is calculated resulting in 281.39 grams.

3 0

4 years ago