The density of an ordinay rock is close to 3 g/cm^3 wihle the density of the paper clips is close to 8 g/cm^3 (the density of steel), then equal apparent volumes (same box) will contain different mass, being of course the mass of the box with paper clips much higher than that of the box with rocks.
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Answer:</h3>
0.89 J/g°C
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Explanation:</h3>
Concept tested: Quantity of heat
We are given;
- Mass of the aluminium sample is 120 g
- Quantity of heat absorbed by aluminium sample is 9612 g
- Change in temperature, ΔT = 115°C - 25°C
= 90°C
We are required to calculate the specific heat capacity;
- We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.
That is;
Q = m × c × ΔT
- Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.
Specific heat capacity, c = Q ÷ mΔT
= 9612 J ÷ (120 g × 90°C)
= 0.89 J/g°C
Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C
Answer:
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon).
Explanation: pls mark brainliest :))
The balanced neutralization reaction here is:
Ca(OH)2 + 2HBr --> 2H2O + CaBr2
Notice that two moles of Her are required to neutralize every one mole of Ca(OH)2. This means that for however many moles of Her reacted, HALF as many moles of Ca(OH)2 reacted as well.
Moles of HBr reacted = 0.75 M x 0.345 L = 0.259 mol
Moles of Ca(OH)2 reacted = 0.259 mol / 2 = 0.130 mol
Concentration of Ca(OH)2 = 0.130 mol / 0.250 L = 0.52 M
We need to find the Ka value of HF. First, set up the balanced chemical equation:
HF ===> H+ + F-
The formula for the Ka value of an acid is
Ka = [concentration of products] / [concentration of reactants]
Ka = [H+] [F-} / [HF]
So if we are given the concentration of the products and reactants, we will be able to solve for the Ka value of HF.
