Answer:
Newton's Second Law
Explanation:
Newton's second law basically states that the acceleration of a body which is produced by a net force is directly proportional to the magnitude of net force applied in the same direction.
This tells us that
F is directly proportional to a
⇒ F= ma
So we can also state from the above equation, that when we have more mass, we need more net force to accelerate it. Here, we are keeping the acceleration constant so we can surely say that force and mass varies directly.
Therefore, we have made good use of Newton's Second Law of motion to arrive at this conclusion.
Answer:
For example, the relative rate of a reaction at 20 seconds will be 1/20 or 0.05 s -1, while the average rate of reaction over the first 20 seconds will be the change in mass over that period, divided by the change in time.
Explanation:
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Answer:
The volume is 59, 3 liters. See the explanation below, please
Explanation:
STP conditions (standard) correspond to 273K of temperature and 1 atm of pressure. These values are used and the volume is calculated, according to the formula:
PV = nRT
1 atm x V= 2, 65 moles x 0, 082 l atm/K mol x 272 K
V= 2, 65 moles x 0, 082 l atm/K mol x 272 K/1 atm = 59, 3 liters
Answer:
O B. Convert the 10 g of NaCl to moles of NaCl.
Explanation:
The formula for finding the molality is m=moles of solute/kg of solvent. The solute for this question is NaCl and the solvent is water.
(10g NaCl)(1 mol NaCl/58.44g NaCl)=0.1711 mol NaCl
58.44 is the molar mass of NaCl
m=0.1711 mol NaCl/2 kg H2O
m=0.085557837
Answer:
HCl(aq) + KOH(aq) ===> H2O(l) + KCl(aq)
Note the stoichiometry of the balanced equations shows us that HCl and KOH react in a 1:1 mole ratio. So, let us find moles of HCl and moles of KOH that are present:
moles HCl = 250.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.06250 moles HCl
moles KOH = 200.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.0800 moles KOH
You can see that there are more moles of KOH than there are of HCl, meaning that KOH is in excess and after neutralizing all of the HCl, the solution will be left with excess KOH making the pH > 7 = BASIC
