Answer: -
15.55 M
35.325 molal
Explanation: -
Let the volume of the solution be 1000 mL.
Density of nitric acid = 1.42 g/ mL
Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid
= 1000 mL x 1.42 g/ mL
= 1420 g.
Percentage of HNO₃ = 69%
Amount of HNO₃ = 
= 979.8 g
Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol
Number of moles of HNO₃ = 
= 15.55 mol
Molarity is defined as number of moles per 1000 mL
We had taken 1000 mL as volume and found it to contain 15.55 moles.
Molarity of HNO₃ = 15.55 M
Mass of water = Total mass of nitric acid solution - mass of nitric acid
= 1420 - 979.8
= 440.2 g
So we see that 440.2 g of water contains 15.55 moles of HNO₃
Molality is defined as number of moles of HNO₃ present per 1000 g of water.
Molality of HNO₃ = 
= 35.325 molal
Answer:
new substances formed, as evidenced by the color changes and bubbles. Some signs of a chemical change are a change in color and the formation of bubbles. The five conditions of chemical change: color chage, formation of a precipitate, formation of a gas, odor change, temperature change.
HOPE THIS HELPED!!!!!!!!!!!!!!XDDDDDDDDDD
In the physical properties, it is mentioned that the element has 4 valence electrons. The elements in the periodic table are arranged such that, the number of valence electrons present in the neutral atoms belonging to a particular group is equal to the group number.
Thus, the unidentified element can be best classified as a nonmetal in period 4
Ans C)
Answer:
7.60 M
Explanation:
Our method to solve this question is to use the definition of molarity (M) concentration which is the number of moles per liter of solution, so for this problem we have
[Cl⁻] = # mol Cl⁻ / Vol
Now the number of moles of Cl⁻ will be sum of Cl in the 1.00 mL 5.4 M solution plus the moles of Cl⁻ in the 0.50 mL 12 M H . Since the volume in liters times the molarity gives us the number of moles we will have previous conversion of volume to liters for units consistency:
1mL x 1 L / 1000 mL = 0.001 L
0.5 mL x 1L/1000 mL = 0.0005 L
[Cl⁻] = 0.001 L x 5.4 mol/L + 0.0005L x 12 mol/L / ( 0.001 L + 00005 L )
= 7.6 M
This is the same as the statement given in the question.
