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AURORKA [14]
3 years ago
7

How is science used by a firefighter

Chemistry
1 answer:
Rufina [12.5K]3 years ago
4 0
Science is applied in all parts of our lives. Firefighters use science by knowing what causes fire and what circumstances are necessary for a fire to occur naturally. for example knowing the chemical composition of hay required in order for it to self ignite on fire helps to take measurements and prevent unwanted fires.
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The Kₐ of carbonic acid is 4.3 x10⁻⁷ . H₂CO₃ ⇄ H⁺ + HCO₃ ⁻ This means that H₂CO₃ is a
sashaice [31]

Answer:

A good hydrogen ion acceptor

8 0
3 years ago
Identify the conjugate base in each pairs
DiKsa [7]

Answer: 1) RCOO^-

2) H_2PO_4^-

3) RNH_2

4) HCO_3^-

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

1) RCOOH\rightarrow RCOO^-+H^+

Here, RCOOH is loosing a proton, thus it is considered as an acid and after losing a proton, it forms RCOO^- which is a conjugate base.

2) H_3PO_4\rightarrow H_2PO_4^-+H^+

Here, H_3PO_4 is loosing a proton, thus it is considered as an acid and after losing a proton, it forms H_2PO_4^- which is a conjugate base.

3) RNH_3^+\rightarrow RNH_2+H^+

Here, RNH_3^+ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms RNH_2 which is a conjugate base.

4) H_2CO_3\rightarrow HCO_3^-+H^+

Here, H_2CO_3 is loosing a proton, thus it is considered as an acid and after losing a proton, it forms HCO_3^- which is a conjugate base.

8 0
3 years ago
Calculate the enthalpy of reaction for 2CO + O2 → 2CO2. given the following bond energies:
kotegsom [21]

Answer : The enthalpy change for the reaction is 1043 kJ/mol.

Explanation :

The given chemical reaction is:

2CO+O_2\rightarrow 2CO_2

As we know that:

The enthalpy change of reaction = E(bonds broken) - E(bonds formed)

\Delta H=[(2\times B.E_{C\equiv O})+(1\times B.E_{O\equiv O})]-[2\times B.E_{C=O}]

Given:

B.E_{C\equiv O} = 1074 kJ/mol

B.E_{O\equiv O} = 499 kJ/mol

B.E_{C=O} = 802 kJ/mol

Now put all the given values in the above expression, we get:

\Delta H=[(2\times 1074kJ/mol)+(1\times 499kJ/mol)]-[2\times 802kJ/mol]

\Delta H=1043kJ/mol

Therefore, the enthalpy change for the reaction is 1043 kJ/mol.

7 0
3 years ago
What percentage of carpet is calcium carbonates?
e-lub [12.9K]

Answer:

the percentage of carpet is calcium carbonates is 12.456734190

5 0
3 years ago
How many valence electrons does an atom of al possess?
Irina18 [472]

Answer:

3

Explanation:

Third element to right

3 0
3 years ago
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