Answer: 5622.6g
Explanation:
Note: Kf for water is 1.86°C/m.
The simple calculation is in the attachment below.
The top number on elements in the periodic table is the atomic number, so when you look at the periodic table you can see that 42 is Mo (Molybdenum), and this is the only element with that atomic number.
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The answer is 20N as when the object is accelerating at 3 m/s2 it as a force of 10N applied to it so to get to 6m/s2 3(x2) = 6 so 10(x2) = 20
Answer:
The athlete should consume about 476 g of carbohydrates per day to maintain a good storage of glycogen in his body
Explanation:
For an athlete who trains daily, 5 to 7 g of carbohydrates per Kg of body weight per day is recommended;
∴ m athlete = (175 Lb)×(0.453592 Kg/Lb) = 79.3787 Kg
⇒ m carbohydrate = (79.3787 Kg)×( 6 g carbohydrate/ Kg.day) = 476.27 g carbohydrate/day
Answer:
d) 8.01 E23 atoms
Explanation:
∴ mass C = 16 g
∴ molar mass C = 12.0107 g/mol
⇒ mol C = (16 g)*(mol/12.0107 g) = 1.332 mol
⇒ atoms C = (1.332 mol)*(6.022 E23 atoms/mol)
⇒ atoms C = 8.02 E23 atoms