Answer:
C
Explanation:
It is most likely non metal with a high electronegativity.
The term that identifies the amount of heat needed to raise the temperature of 1 gram of a substance by 1 degree Celsius is Specific heat capacity.
<h3>
What is specific heat capacity?</h3>
Specific heat capacity is the quantity of heat needed to raise the temperature of 1 gram of a substance by 1 degree Celsius. The unit of specific heat capacity is joule per gram per degree Celsius (J/g⁰C).
<h3>Difference between
heat capacity and
specific heat capacity</h3>
Specific heat capacity is heat required to raise the temperature of a unit mass of a substance while heat capacity is the quantity of heat required to raise the temperature an entire mass of a substance.
Heat capacity is measure in Joules (J) while specific heat capacity is measured in joule per gram per degree Celsius (J/g⁰C).
Thus, the term that identifies the amount of heat needed to raise the temperature of 1 gram of a substance by 1 degree Celsius is Specific heat capacity.
Learn more about specific heat capacity here: brainly.com/question/16559442
Answer: The molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
Explanation:
Molarity is the number of moles of a substance present in liter of a solution.
And, moles is the mass of a substance divided by its molar mass.
(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.
![Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C%3D%20%5Cfrac%7B28.5%20g%7D%7B46%20g%2Fmol%7D%5C%5C%3D%200.619%20mol)
Now, molarity of ethanol solution is as follows.
![Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M](https://tex.z-dn.net/?f=Molarity%20%3D%20%5Cfrac%7Bmoles%7D%7BVolume%20%28in%20L%29%7D%5C%5C%3D%20%5Cfrac%7B0.619%20mol%7D%7B4.50%20%5Ctimes%2010%5E%7B2%7D%20%5Ctimes%2010%5E%7B-3%7DL%7D%5C%5C%3D%201.38%20M)
(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.
![Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C%3D%20%5Cfrac%7B21.6%20g%7D%7B342.3%20g%2Fmol%7D%5C%5C%3D%200.063%20mol)
Now, molarity of sucrose solution is as follows.
![Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M](https://tex.z-dn.net/?f=Molarity%20%3D%20%5Cfrac%7Bmoles%7D%7BVolume%20%28in%20L%29%7D%5C%5C%3D%20%5Cfrac%7B0.063%20mol%7D%7B0.067%20L%7D%20%20%281%20mL%20%3D%200.001%20L%29%5C%5C%3D%200.94%20M)
(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.
![Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C%3D%20%5Cfrac%7B6.65%20g%7D%7B58.44%20g%2Fmol%7D%5C%5C%3D%200.114%20mol)
Now, molarity of sodium chloride solution is as follows.
![Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M](https://tex.z-dn.net/?f=Molarity%20%3D%20%5Cfrac%7Bmoles%7D%7BVolume%20%28in%20L%29%7D%5C%5C%3D%20%5Cfrac%7B0.114%20mol%7D%7B0.0962%20L%7D%5C%5C%3D%201.182%20M)
Thus, we can conclude that the molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
To work this out you do 400÷20=20