Which substance can be decomposed by chemical means?

The tendency to combust is a chemical property. O True O False

erastova [34]

Answer:

The correct option is;

True

Explanation:

Combustion is a chemical reaction between components of a combustion reaction which are;

1) A fuel

2) An oxidizing agent

The oxidizing agent or oxidant in a combustion reaction consist mainly of oxygen.

A combustion reaction is an oxidation and reduction (redox) heat evolving (exothermic) reaction that takes place at high temperatures and produces smoke which consists of gaseous mixture of chemicals and sometimes very small solid particles

An exothermic reaction can be classed either as rapid, spontaneous, or explosive

A spontaneous exothermic reaction is one that begins on its own in atmospheric conditions and room temperature as the heat required to begin the reaction is produced by the initial oxidation of the reacting fuel

Example of a substance that undergoes spontaneous oxidation reaction is white phosphorus.

4 0

3 years ago

What happens when elements combine to form compounds?

storchak [24]

When the elements combine form compounds, they do it to become stable. They give and taker the outside electrons or by sharing outer ring electrons.

7 0

3 years ago

The following Lewis diagram represents the valence electron configuration of a main-group element. This element is in group 2A A

Paraphin [41]

Answer: The symbol of the ion formed is $Ca^{2+}$

Explanation:

An ion is formed when a neutral atom looses or gains electrons.

When an atom looses electrons, it results in the formation of positive ion known as cation.
When an atom gains electrons, it results in the formation of negative ion known as anion.

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

The element present in Group 2-A and in period 4 is Calcium (Ca)

Electronic configuration of Ca atom: $1s^22s^22p^63s^23p^64s^2$

This atom will loose 2 electrons to attain stable electronic configuration similar to Argon element (noble gas)

The electronic configration of $Ca^{2+}\text{ ion}=1s^22s^22p^63s^23p^6$

Hence, the symbol of the ion formed is $Ca^{2+}$

4 0

3 years ago

1(30%). A thin plate black body, insulated at the bottom is placed faced up in a room with wall temperatures of 30 C. Air at 0 C

ch4aika [34]

Explanation:

The given data is as follows.

Heat transfer coefficient (h) = 12 $W/m^{2}.C$

Plate temperature ( $T_{1}$ ) = $30 ^{o}C$ = 303 K

Steady state temperature ( $T_{2}$ ) = ?

Hence, formula applied for steady state is as follows.

$(h \times A \times \Delta T)_{air}$ = $\sigma \times A \times (T^{4}_{1} - T^{4}_{2})_{plate}$

Putting the given values into the above formula as follows.

$(h \times A \times \Delta T)_{air}$ = $\sigma \times A \times (T^{4}_{1} - T^{4}_{2})_{plate}$

$12 \times (T_{2} - 273)$ = $5.67 \times 10^{-8} \times [(30 + 273)^{4} - T^{4}_{2}]$

$T_{2}$ = 282.66 K

= (282.66 -273) $^{o}C$

= 9.66 $^{o}C$

Thus, we can conclude that the steady state temperature will be 9.66 $^{o}C$ .

3 0

3 years ago

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

Answer: The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago