Answer:
It is all about the tilt of the Earth's axis. Many people believe that the temperature changes because the Earth is closer to the sun in summer and farther from the sun in winter. In fact, the Earth is farthest from the sun in July and is closest to the sun in January!
Half-life refers to the time taken for half of the atoms in a radioactive substance to decay. Therefore in 1 half-life the 80 mg of substance will reduce to 40 mg. In another half-life it will reduce from 40 mg to 20 mg, then in a third half-life from 20 mg to 10 mg. Therefore the half-life is equal to the total time taken to decay from 80 mg to 10 mg divided by 3:
t(1/2) = 36 hours / 3 = 12 hours.
Answer:
from the pic:
since molar mass is 118.084 g/mol;
It is an aldehyde with structure ( second pic )
<h3>
Answer:</h3>
Ag⁺(aq) +Cl⁻(aq) → AgCl(s)
<h3>
Explanation:</h3>
The questions requires we write the net ionic equation for the reaction between aqueous potassium chloride and aqueous silver nitrate.
<h3>Step 1: Writing a balanced equation for the reaction.</h3>
- The balanced equation for the reaction between aqueous potassium chloride and aqueous silver nitrate will be given by;
KCl(aq) + AgNO₃(aq) → KNO₃(aq) +AgCl(s)
- AgCl is the precipitate formed by the reaction.
<h3>Step 2: Write the complete ionic equation.</h3>
- The complete ionic equation for the reaction is given by showing all the ions involved in the reaction.
K⁺(aq)Cl⁻(aq) + Ag⁺(aq)NO₃⁻(aq) → K⁺(aq)NO₃⁻(aq) +AgCl(s)
- Only ionic compounds are split into ions.
<h3>Step 3: Write the net ionic equation for the reaction.</h3>
- The net ionic equation for a reactions only the ions that fully participated in the reaction and omits the ions that did not participate in the reaction.
- The ions that are not involved directly in the reaction are known as spectator ions and are not included while writing net ionic equation.
Ag⁺(aq) +Cl⁻(aq) → AgCl(s)
Answer:
The new volume after the temperature reduced to -100 °C is 0.894 L
Explanation:
Step 1: Data given
Volume of nitrogen gas = 1.55 L
Temperature = 27.0 °C = 300 K
The temperature reduces to -100 °C = 173 K
The pressure stays constant
Step 2: Calculate the new volume
V1/T1 = V2/T2
⇒with V1 = the initial volume of the gas = 1.55 L
⇒with T1 = the initial temperature = 300 K
⇒with V2 = the new volume = TO BE DETERMINED
⇒with T2 = the reduced temperature = 173 K
1.55 L / 300 K = V2 / 173 K
V2 = (1.55L /300K) * 173 K
V2 = 0.894 L
The new volume after the temperature reduced to -100 °C is 0.894 L
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