I would choose C bc if u look at it closely u can notice it goes up steadily and then BANG it decreases a lot
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
There are:
19.8g of nuts (90x0.22=19.8)
31.5g of granola (90x0.35=31.5)
16.2g of dried fruit (90x0.18=16.2)
22.5g of chocolate chips (90x0.25=22.5)
The heat used in phase changes is calculated by multiplying the mass of the substance by the energy of the phase change. In this case, for liquid to boil, we would find total heat by multiply the mass of liquid by the latent heat of vaporization (Hvap). If we are instead given the Hvap and the total heat of 1 kJ, we would divide 1 kJ by the Hvap (which is usually in kJ/kg) to get the mass of liquid boiled (in kg).
Answer: There are 
atoms of hydrogen are present in 40g of urea, 
.
Explanation:
Given: Mass of urea = 40 g
Number of moles is the mass of substance divided by its molar mass.
First, moles of urea (molar mass = 60 g/mol) are calculated as follows.
According to the mole concept, 1 mole of every substance contains 
atoms.
So, the number of atoms present in 0.67 moles are as follows.
In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.
Thus, we can conclude that there are atoms of hydrogen are present in 40g of urea, .
