Answer:
2.74 M
Explanation:
Given data:
Mass of sodium chloride = 80.0 g
Volume of water = 500.0 mL
Molarity of solution = ?
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Now we will convert the mL into L.
500.0 mL× 1 L /1000 mL = 0.5 L
In next step we will calculate the number of moles of sodium chloride.
Number of moles = mass/molar mass
Number of moles = 80.0 g/ 58.4 g/mol
Number of moles = 1.37 mol
Molarity:
M = 1.37 mol/ 0.5 L
M = 2.74 M
Answer:
333N
Explanation:
Once we have the object's mass, we can find the weight by multiplying it by the gravitational force, which it is subject to. Being that Mars has a gravitational force of 3.7m/s2, we multiply the object's mass by this quantity to calculate an object's weight on mars.
So an object or person on Mars would weigh 37.83% its weight on earth. Therefore, a person would be much lighter on mars. Conversely, a person is 62.17% heavier on earth than on Mars.
Answer:
Technician B.
Explanation:
The claim of technician B that the PCM will add fuel if the IAT indicates that the incoming air temperature is cold is correct.
Answer is: 56 gallons of
70% antifreeze and 84 gallons of 95% antifreeze.
ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.
ω₂ = 95% ÷ 100% = 0.95.
ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.
V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.
V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.
0.70 · (140 gal -
V₂) + 0.95 · V₂ = 0.85 · 140 gal.
98 gal - 0.7V₂ + 0.95V₂ = 119 gal.
0.25V₂ = 21 gal.
V₂ = 21 gal ÷ 0.25.
V₂ = 84 gal.
V₁ = 140 gal - 84 gal.
V₁ = 56 gal.
Answer:
The pH of the buffer solution is 4.60.
Explanation:
Concentration of acid = ![[HC_2H_3O_2]=0.225 M](https://tex.z-dn.net/?f=%5BHC_2H_3O_2%5D%3D0.225%20M)
Concentration of salt = ![[KC_2H_3O_2]=0.162 M](https://tex.z-dn.net/?f=%5BKC_2H_3O_2%5D%3D0.162%20M)
Dissociation constant = 
The pH of the buffer can be determined by Henderson-Hasselbalch equation:
![pH=pK_a+\log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
![pH=-\log[1.8 \times 10^{-5}]+\log\frac{0.162 M}{0.225 M}](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B1.8%20%5Ctimes%2010%5E%7B-5%7D%5D%2B%5Clog%5Cfrac%7B0.162%20M%7D%7B0.225%20M%7D)
pH = 4.60
The pH of the buffer solution is 4.60.
