Answer:
58.44 g/mol The Molarity of this concentration is 0.154 molar
Explanation:
the molar mass of NaCl is 58.44 g/mol,
0.9 % is the same thing as 0.9g of NaCl , so this means that 100 ml's of physiological saline contains 0.9 g of NaCl. One liter of physiological saline must contain 9 g of NaCl. We can determine the molarity of a physiological saline solution by dividing 9 g by 58 g... since we have 9 g of NaCl in a liter of physiological saline, but we have 58 grams of NaCl in a mole of NaCl. When we divide 9 g by 58 g, we find that physiological saline contains 0.154 moles of NaCl per liter. That means that physiological saline (0.9% NaCl) has a molarity of 0.154 molar. We can either express this as 0.154 M or 154 millimolar (154 mM).
load every electric circuit,regardless of where it is or how large or small, has four basic parts: an energy source (ac or dc),a conductor (wire), an electrical load (device), and at least one controller(switch)
Answer: new depth will be 3.462m and the water elevation will be 0.462m.
The maximum contraction will be achieved in width 0<w<3
Explanation:detailed calculation and explanation is shown in the image below
Answer:
<u>Option-(A)</u>
Explanation:
<u>Typical applications for the high carbon steels includes the following;</u>
It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.
Answer:
c. less than 60 mi/h
Explanation:
To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.
Total Distance Traveled = S = 100 mi + 100 mi
S = 200 mi
Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.
Total Time = t = Time from A to B + Time from B to C
t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)
t = 2 h + 1.43 h
t = 3.43 h
Now, the average speed of bus will be given as:
Average Speed = V = S/t
V = 200 mi/3.43 h
<u>V = 58.33 mi/h</u>
It is clear from this answer that the correct option is:
<u>c. less than 60 mi/h</u>