# Posts by Jayd

Total # Posts: 51

**Pi**

Sorry it posted twice

*February 13, 2017*

**Pi**

I got the answer of 22 So how would i put that

*February 13, 2017*

**Pi**

I got the answer of 22 So how would i put that

*February 13, 2017*

**Pi**

How do u describe how to check an algebra question??? I need help ASAP

*February 13, 2017*

**math**

It is 22 just describe it how to check it

*January 4, 2011*

**Physics**

A 9 kg lead brick falls from a height of 1.8 m. a) Find the momentum as it reaches the ground. b) What impulse is needed to bring the brick to rest? c) The brick falls onto a carpet 2.0 cm thick. Assuming the force stopping it is constant, find the average force the carpet ...

*November 7, 2007*

**Physics**

12 - -32.1 = 44.1 I did not think it was this easy. I plugged the data into like four formulas trying to come up with an answer.

*September 4, 2007*

**Physics**

Kyle is flying a helicopter and it is rising at 4.0 m/s when he releases a bag. After 3.0s a) What is the bag's velocity? -25.4 b) How far has the bag fallen? -32.1 c) How far below the helicopter is the bag? Helicopter: a = -9.80 m/s^2 V = 4.0 m/s t = 3.0 s d = ?

*September 4, 2007*

**Physics**

When I solve for a I get the .061 and this is incorrect for the constant acceleration. I don't know what I am going wrong.

*September 4, 2007*

**Physics**

When I solve for a I get the .061 and this is incorrect for the constant acceleration. I don't know what I am going wrong.

*September 4, 2007*

**Physics**

A car moving with a constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its velocity as it passes the second point is 17 m/s. a) What was the speed at the first point? b) What is the constant acceleration? c) How far behind the first point was the...

*September 4, 2007*

**Physics**

35 - (.9)(-10.0) = 44 V = 29.664 and this is incorrect Do I use the acceleration for the velocity or do I figure out the velocity from the informatin given?

*September 4, 2007*

**Physics**

What is the maximum speed at which a car could be moving and not hit a barrier 35.0 m ahead if the average acceleration during braking is -10.0 m/s2 and it takes the driver 0.90 s before he applies the brakes?

*September 4, 2007*

**Physics**

d = 165^2/(2 x 32.0) d = 425.390

*September 4, 2007*

**Physics**

acceleration = -32.0 m/s vf = 165 m/s vi = 0 d = ? Vf^2 = Vi^2 + 2ad d = Vf^2 - Vf^2/2a

*September 4, 2007*

**Physics**

A rocket traveling at 165 m/s is accelerated at a rate of -32.0 m/s. a) How long will it take before the instantaneous speed is 0 m/s? It takes 5.156 s. b) How far will it travel during this time?

*September 4, 2007*

**Physics**

Car A is traveling at 84 km/h while Car B is traveling at 58 km/h. What is the relative velocity of Car A to Car B if they both are traveling in the same direction? What is the relative velocity of Car A to Car B if they are headed towards each other?

*August 26, 2007*

**Physics**

Velocity = displacement/time Relative velocity = velocity with which a body approaches or recedes from another body Would you add the speeds of both cars in order to find the relative velocity between the two when they are traveling in the same direction because the direction ...

*August 25, 2007*

**Physics**

Car A is traveling at 84 km/h while Car B is traveling at 58 km/h. What is the relative velocity of Car A to Car B if they both are traveling in the same direction? What is the relative velocity of Car A to Car B if they are headed towards each other?

*August 25, 2007*

**Physics**

The answer would be 52 km/h.

*August 25, 2007*

**Physics**

This is the answer that I got when I worked it, but the webassign homework that I have marked the answer incorrect. I thought that maybe I was doing something wrong or missing an important part of the problem, thus causing my answer to be wrong.

*August 25, 2007*

**Physics**

You drive your car from home at an average spped of 78km/h for 2 h. Halfway to your destination, you develop some engine problems, and for 4 h you nurse the car the rest of the way. What is your average velocity for the entire trip?

*August 25, 2007*

**Physics**

The average distance between Earth and the sun is 1.50e8 km. Calculate the average speed, in km/h, of Earth assuming a circular path about the sun. Use the equations v = 2(pi)r/T. You know that, for the Earth, T = 365.25 days. Convert that to hours (by multiplying by 24 h/day...

*August 16, 2007*

**Phyics**

Water drips from a faucet into a flask at the rate of two drops every 2 seconds. A cubic centimeter (cm3) contains 20 drops. What volume of water, in cubic decimeters (dm3), will be collected in 3 hours? 3600 seconds in one hour 10800 seconds in three hours therefore there are...

*August 16, 2007*

**Phyics**

The radius of Earth is 6.37 103 km. Find the speed, in km/h, resulting from the rotation of Earth, of a person standing on the equator. v=2(pi)(r)/T v=2(pi)(6.37e3)/24 v=1.67e3 km/h how would you convert this to m/s. it does not seem to be working for me and I cannot find my ...

*August 16, 2007*

**Phyics**

The average distance between Earth and the sun is 1.50e8 km. Calculate the average speed, in km/h, of Earth assuming a circular path about the sun. Use the equations v = 2r/T. Please ignore this question and view the question I have a few spaces above. The equation was not ...

*August 16, 2007*

**Chemistry II**

The value of ^G for the following reaction is -5490 kJ. Use this value and data to calculate the standard free energy of formation for C4H10 (g). 2 C4H10 (g) + 13 O2 (g) --> 8 CO2 (g) + 10 H2o (l) Do you know the heats of formation for CO2 and H2O? Wouldn't delta G/2, ...

*May 23, 2007*

**Chemistry II**

The value of ^G for the following reaction is -5490 kJ. Use this value and data to calculate the standard free energy of formation for C4H10 (g). 2 C4H10 (g) + 13 O2 (g) --> 8 CO2 (g) + 10 H2o (l)

*May 22, 2007*

**Chemistry II**

Are mixing and separation increases in entropy?

*May 22, 2007*

**Chemistry II**

What is the ^S for NaCl (l)?

*May 22, 2007*

**Chemistry II**

The value of ^G for the following reaction is -5490 kJ. Use this value and data to calculate the standard free energy of formation for C4H10 (g). 2 C4H10 (g) + 13 O2 (g) --> 8 CO2 (g) + 10 H2o (l)

*May 22, 2007*

**Chemistry II**

I will be happy to critique your thinking on this. First, calculate the moles of H2 used. 2.0e8 L H2/22.4 L H2 = 8.9e6 mol H2 For every 2 mol H2 used -572 kJ of energy is produced -572 kJ/2 mol H2 = -2.86e2 kJ/mol I have a balanced equation with this problem. It is: 2 H2 + O2...

*May 20, 2007*

**Chemistry II**

The enthalpy of neutralizatino for the reaction of a strong acid with a strong base is -56 kJ/mol of water produced. How much energy will be released when 230.0 mL of 0.400 M HCl is mixedwith 150.5 mL of 0.500 M NaOH? How do I figure out how much water is produced in the ...

*May 20, 2007*

**Chemistry II**

The total volume of hydrogen gas needed to fill the Hidenbrug was 2.0e8 L at 1.0 atm and 25 degrees Celsius. How much heat was evolved when the Hidenburg exploded, assuming all the hydrogen reacted? 2.0e8 L H2/22.4 L H2 = 8.9e6 mol H2 For every 2 mol H2 used -572 kJ of energy ...

*May 20, 2007*

**Chemistry II**

The total volume of hydrogen gas needed to fill the Hidenbrug was 2.0e8 L at 1.0 atm and 25 degrees Celsius. How much heat was evolved when the Hidenburg exploded, assuming all the hydrogen reacted? I will be happy to critique your thinking on this. First, calculate the moles ...

*May 20, 2007*

**Chemistry II**

The enthalpy of neutralizatino for the reaction of a strong acid with a strong base is -56 kJ/mol of water produced. How much energy will be released when 230.0 mL of 0.400 M HCl is mixedwith 150.5 mL of 0.500 M NaOH? How much water is produced in the reaction? How do I figure...

*May 20, 2007*

**Chemistry II**

Calculate the change in heat: 2 Na(s) + 2 H2O(l) > 2 NaOH(aq) + H2(g) So far this is what I have: 2 H20 --> 2 H2 + O2 2(285.83) kJ 2 Na + O2 + H2 --> 2 NaOH 2(-426.73) kJ 2 Na + 2 H2O --> 2 NaOH + H2 -281.8 kJ If this correct?? yes, correct, as best as I can tell.

*May 20, 2007*

**Chemistry II**

C4H4 (g) + 2 H2 (g) --> C4H8 (g) Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for C4H4 (-2341 kJ/mol), C4H8 (-2755 kJ...

*May 20, 2007*

**Chemistry II**

The total volume of hydrogen gas needed to fill the Hidenbrug was 2.0e8 L at 1.0 atm and 25 degrees Celsius. How much heat was evolved when the Hidenburg exploded, assuming all the hydrogen reacted?

*May 20, 2007*

**Chemsitry II**

The partial pressure of CH4(g) is 0.185 atm and that of O2(g) is 0.300 atm in a mixture of the two gases. a) What is the mole fraction of each gas in the mixture? b) If the mixture occupies a volume of 11.5 L at 65 degress C, calculate the total number of moles of gas in the ...

*May 10, 2007*

**Chemistry II**

If 36.0 mL of 7.0e-4 M HClO4 is added to 19.5 mL of 8.2e-4 M LiOH, what is the pH of the solution? Write the equation. Calculate mols HCLO4. Calculate mols LiOH. Determine how much of which reagent remains after the reaction. pH = -log(H^+) post your work if you get stuck. ...

*April 25, 2007*

**Chemistry II**

A 12.00 mL sample of sulfuric acid from an automobile battery requires 34.62 mL of 2.42 M sodium hydroxide solution for complete neutralization. What is the molarity of the sulfuric acid? H2SO4 + 2NaOH ==> Na2SO4 + 2H2O molarity NaOH x liters NaOH = mols NaOH. mols H2SO4 = ...

*April 25, 2007*

**Chemistry II**

What volume of .0500 M Ba(OH)2 will react completely with 21.00 mL of .500 M HCl? Same type problem. Post your work if you get stuck. 2HCl + Ba(OH)2 --> 2HOH + BaCl2 (.0500 mol HCl / 1000 mL) (21.00 mL) (1 mol Ba(OH)2 / 2 mol HCl) (1000 mL / .0500 mol Ba(OH)2) = 1.05e2 mL ...

*April 25, 2007*

**Chemistry II**

Hydrochloric acid (77.0 mL of .267 M) is added to 236.0 mL of .0730 M Ba(OH)2 solution. What is the concentration of the excess H+ or OH- ions left in the solution? Done two of these. They are all alike. Use the template from the previous problem. Post your work if you get ...

*April 25, 2007*

**Chemistry II**

What volume of .0500 M calcium hydroxide is required to neutralize 38.50 mL of .0400 M nitric acid? 1. Write the equation. 2. M HNO3 x L HNO3 = mols HNO3. 3. Use the equation to convert mols HNO3 to mols Ca(OH)2. 4. Now use M x L = mols to calculate volume (L) Ca(OH)2. I ...

*April 25, 2007*

**Chemistry II**

What volume of .100 M NaOH will react completely with 21.00 mL of .500 M HCl? Vb*ConcBase= Vacid*ConcAcid Solve for Vb. 21.00 mL x .500 M HCl = 1.05e1 1.05e1 / .100 M NaOH = 1.05e2 right.

*April 25, 2007*

**Chemistry II**

If 15.0 mL of 1.2e-4 mol/L HI is added to 26.0 mL of 9.2e-4 mol/L HI, what is the pH of the solution? M x L = mols. Calculate mols for solution 1, do the same for solution 2, add mols together and divide by total liters. That will be the new molarity. Then pH = -log(Molarity)

*April 24, 2007*

**Chemistry II**

A stock solution containing manganese ions was prepared by dissolving 1.542 g pure manganese metal in nitric acid and diluting to a final volume of 1.000L. Calculate the concentration of the stock solution. mols Mn = g/atomic mass Mn. M = mols/L 1.542 g / 54.9 amu Mn = .028 ...

*April 16, 2007*

**Chemistry II**

A solution of ethanol in water is prepared by dissolving 61.0 mL of ethanol (density = 0.79g/mL) in enough water to make 280.0 mL of solution. What is the molarity of the ethanol in this solution? definition of molarity = #mols/L. mass ethanol = m = volume x density = 61.0 mL ...

*April 16, 2007*

**Chemistry II**

Calculate the sodium ion concentration when 45.0 mL of 2.5 M sodium cabonate is added to 30.0 mL of 1.0 M sodium bicarbonate. mols Na2CO3 = M x L. Mols Na^+ = twice that. mols NaHCO3 = M x L. Mols Na^+ = 1 x that. Total Na = sum of the two.

*April 16, 2007*

**Chemistry II**

How much of solid potassium chromate whould you use to prepare 2.00 L of 0.400 M potassium chromate? M x L = mols g/molar mass = mols OR M x L = g/molar mass solve for g. .400 mol/L x 2.00 = g/194 g = 155.2 Thanks!!

*April 16, 2007*

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