The statement that defines the specific heat capacity for a given sample is the quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.
<h3>What is specific heat capacity?</h3>
Specific heat capacity is the of heat to increase the temperature per unit mass.
The formula to calculate the specific heat is Q = mct.
The options are attached here:
- The temperature of a given sample is 1 %.
- The temperature that a given sample can withstand.
- The quantity of heat that is required to raise the sample's temperature by 1 °C1 °C (Kelvin).
- The quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.
Thus, the correct option is 4. The quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.
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Answer:
The required volume is 1.6 x 10³mL.
Explanation:
When we want to prepare a dilute solution from a concentrated one, we can use the dilution rule to find out the required volume to dilute. This rule states:
C₁ . V₁ = C₂ . V₂
where,
C₁ and V₁ are the concentration and volume of the concentrated solution
C₂ and V₂ are the concentration and volume of the dilute solution
In this case, we want to find out V₁:
C₁ . V₁ = C₂ . V₂
Answer:
A one-step mechanism involving a transition state that has a carbon partially bonded to both chlorine and oxygen
Explanation:
The compound CH3Cl is methyl chloride. This is a nucleophilic substitution reaction that proceeds by an SN2 mechanism. The SN2 mechanism is a concerted reaction mechanism. This means that the departure of the leaving group is assisted by the incoming nucleophile. The both species are partially bonded to opposite sides of the carbon atom in the transition state.
Recall that an SN2 reaction is driven by the attraction between the negative charge of the nucleophile (OH^-) and the positive charge of the electrophile (the partial positive charge on the carbon atom bearing the chlorine leaving group).
Explanation:
This is correct!
Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.
An example is;
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)
The ions; Na+, NO3−(aq) would be cancelled out to give;
Cl−(aq) + Ag+(aq) → AgCl(s)
