Answer:
Mass = 0.139 kg
Explanation:
There is a formula in terms of force, acceleration and mass which is:
Force = mass × acceleration
Put the values into the formula.
5 = m × 36
m = 5 ÷ 36
<u>m = 0.139 kg</u>
Answer #1 is "there is 2.5 grams of solute in every 100 g of solution."
We calculate for 2.5% by mass solution by dividing the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2 is "that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution."
We weigh out 2.5 grams of solute and then add 97.5 grams of solvent to make a total of 100 gram solution, that is,
mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
= 2.5g solute / 100g solution
Answer#3 is "a solution mass of 1 kg is 10 times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain 25 grams of ki."
We multiply 10 to each mass so that 100 grams becomes 1000grams since 1000 grams is equal to 1 kg:
mass of solute / mass of solution = 2.5g*10/[(2.5g*10) + (97.5g*10)]
= 25g solute/(25g solute + 975g solvent)
= 25g solute/1000g solution
= 25g solute/1kg solution
Base isolation is a method of earthquake resistant engineering where a structure is separated from the ground by isolation units that reduces the impact of seismic vibration energy. So, the last answer choice would be the best definition.
NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)
You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)
So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT
n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)
n = 0.0385 moles H2
moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required
mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
Brainliest Please :)
