Answer:
Explanation:
Did you mean: V = d/t a = (V - Vit Average = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
Showing results for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = "-9.81" m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
Search instead for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?
A general equation for a combustion reaction would be expressed as follows:
CxHy + (x+y/2)O2 = xCO2 + y/2H2O
Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.
moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C
moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H
Then, we divide the smallest amount to the each mole of the atoms. We do as follows:
C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67
Then we multiply a number in order to obtain a whole number ratio between the atoms.
1 CH2.67
2 C2H5.34
3 C3H8 <-------- empirical formula
There are multiple meanings for a base. A base can be a substance that accepts hydrogen ions, or it could be something that is not acidic, in other words meaning its pH is between 7 and 14.
Molarity= mol/ liters
since the molarity is given, we can assume that we have 1.0 Liters of solution
15.6 M= mol/ 1 liters---> this means that we have 15.6 moles of HNO3
we need to convert these moles to grams using the molar mass of HNO3
molar mass HNO3= 1.01 + 14.0 + (3 X 16.0)= 63.01 g/mol
15.6 mol HNO3 (63.01 g/ mol)= 983 grams HNO3
now we have to determine the grams of solution using the assumption of 1 liters of solution and the density
1 liters= 1000 mL
1000 mL (1.41 g/ ml)= 1410 grams solution
mass percent= mass of solute/ mass of solution x 100
mass percent= 63.01/ 1410 x 100= 4.47 %
