To measure weight of a item
Specific heat is the quantity of heat required to change the temperature of 1 gram of a substance by 1 degree Celsius. It is the amount per unit mass that is required to raise the temperature by one degree Celsius. Every substance has its own specific heat and each has its own distinct value. The units of specific heat are joules per gram-degree Celsius (J/f C) and sometimes J/Kg K may also be used.
Interaction occurs when light goes into an object as heat energy is Absorbed light.
Absorbed light makes an object dark. Absorption of light is when the light is absorbed by the object and converts it into heat energy. The objects converts the wavelength of light into the larger wavelength of heat. this makes the object warmer and produced energy. Absorption depends on the frequency of light that is transmitted. The wavelength of light defined as the distance between the the two successive troughs of the the light wave.
Therefore, Interaction occurs when light goes into an object as heat energy is Absorbed light.
To learn more about Absorbed light here
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Answer:
1.2×10² mmole of Na₂S₂O₃
Explanation:
From the question given above, the following data were obtained:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:
Molarity = mole /Volume
With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity = mole /Volume
0.2 = Mole of Na₂S₂O₃ / 0.6
Cross multiply
Mole of Na₂S₂O₃ = 0.2 × 0.6
Mole of Na₂S₂O₃ = 0.12 mole
Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:
1 mole = 1000 mmol
Therefore,
0.12 mole = 0.12 mole × 1000 mmol / 1 mole
0.12 mole = 120 = 1.2×10² mmole
Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃
Answer:
Explanation:
Hello,
In this case, the described chemical reaction is:
Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:
Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:
Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:
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