Check my answers - If a reaction vessel initially contains an N2O4 concentration of 4.50×10−2M at 500 K...?
what are the equilibrium concentrations of N2O4 and NO2 at 500 K for the following reaction, Kc = 0.513 at 500 K. N2O4(g)⇌2NO2(g) I got my quadratic answers, but they both bring up a negative when I solve for x. Here's what I did: 0.513 = 2x^2/ (4.50x10^-2) Solved that into my quadratic of: 4x^2 + 0.513 - 0.0231 After plugging it into the quadratic eqn I got: +0.353 and -.1636 When I solve for X, ((4.5x10^-2 - x) and (2x)) I don't come up with a positive outcome for both of these. Can anyone please help?
Check my work? CALCULUS: Area Under the Curve:: Distance Problem from Velocity and Time...?
Check my work? CALCULUS: Area Under the Curve:: Distance Problem from Velocity and Time...? The velocity graph of a car accelerating from rest to a speed of 90 km/h over a period of 30 seconds is shown. Estimate the distance, d traveled during this period. (Use M6 to get the most precise estimate. Round the answer to two decimal places.) The graph: http://www.webassign.net/scalcet/5-1-016alt.gif where a=30 I used M_6= ∑_(i=1)^6▒〖f(x_i)(∆x)〗 Found ∆x to be 5 seconds, converted seconds into hours (I plugged in 5/3600 to get a more accurate result) and then using the rest of the formula I got ∆x(a/4) + ∆x(7a/4) + ∆x(9a/4) +∆x(5a/2) +∆x(3a) [[They want rectangles, by the way]] My end result was 0.41km, is this right?
Easy physics question, please check my answer?
4. Suppose you are walking home after school. The distance from school to your home is five kilometers. On foot, you can get home in 25 minutes. However, if you rode a bicycle, you could get home in 10 minutes. How much faster would you travel on your bicycle than walking? i got .3 km/ min , is this right? .3km/minute is this right? 5 km 5km
How can I check my visit visa to see if it is valid or not?
Every visa I have in my passports includes either an expiry date or a validity date and validity period. If the later, simply add the number of days to the validity date to get the expiry date.Be aware that in some countries that use different date system (like Saudi Arabia) can have different number of days in a month to the Georgian calendar resulting in mix ups on the expiry date as many people (Saudi govt as well) just count three months for a 90 day visa - I’ve known visitors arrested on the way out being accused of overstaying the visa as a result of these issues. If everyone actually counted the days rather than months, it wouldn’t be an issue.
How do I round my answer to the nearest square kilometer?
It's just the same as with any number. 1799.9 is 1800 to the nearest unit, whatever units you are using.
A plane flying with a constant speed of 4 km/min...calc problem, please help?
The position vector at time zero is r0 = {0,9} The velocity vector is v = 4*{Cos[25], Sin[25]} Therefore, the position vector from the radar as a function of time is r = r0 + v*t = {0,9} + 4*t*{Cos[25],Sin[25]} = {4*t*Cos[25], 9 + 4*t*Sin[25]} = {x, y} The magnitude of r |r| = Sqrt[x^2 + y^2] = Sqrt[(4*t*Cos[25])^2 + (9 + 4*t*Sin[25])^2] =Sqrt[81 + 16 t^2 + 72 t Sin[25]] Now take the derivative with respect to t d|r|/dt = (4 (4 t + 9 Sin[25 °]))/Sqrt[81 + 8 t (2 t + 9 Sin[25 °])] Evaluating this at t = 5 d|r|/dt = 3.784 km/min
Constant Acceleration problem. Physics. Please check my answer.?
I think you are using the right formula for a). The V0... confused me for a second, but i think your formula is v = u + at. v = final velocity, u is initial velocity. As v = 0 (because it has stopped), and a and u are given: 0 = 100 + -5t -100 = -5t 5t = 100 t = 20s b) For this i would turn the 0.5 mile into kilometres. To do this you need to know that 1 mile = 8/5 kilometres, or 1 mile = 1.6 kilometres. (The 8/5 is useful, remember it). Therefore if 1 mile = 1.6 km, then 0.5 miles = 0.8 km. 0.8 km is 800 metres. Therefore you can use another equation of motion. I would use v^2 = u^2 + 2as. As v is again 0, u is 100 m/s, and a is -5 m/s^2: 0^2 = 100^2 + 2 x -5 x s 10000 = 10s s = 1000m So it cannot land on the island because it would require 1000m to land (or 1km) and it only has 800m (or 0.8km). Hope that is useful.
Why did the Yamaha service center tell me to change my chain sprocket if the bike has run only about 6,555 km?
That seems soon. But they could be right, ask a few questions.Have you cleaned and lubricated the chain as per the manual. Probably every 4–500 miles but more frequently in the wet or dusty or muddy environments?Have you checked and adjusted free play when you cleaned the chain?Has the chain adjustment mechanism still got lots of adjustment left on it?Are the teeth on the sprockets still even and blunt/well rounded?If you can answer yes to all of these the sevice center may be ripping you off.If the answer to any of these is no then you just may have prematurely worn out a chain and sprockets with poor or non existent maintenance.
Help with chemistry Homework?
A small speck of carbon the size of a pinhead contains about 5×10^19 atoms, the diameter of a carbon atom is 1.5x10^-10 , and the circumference of the earth at the equator is 40,075 km. How many times around the earth would the atoms from this speck of carbon extend if they were laid side by side? The answer I got was 187.1490954. using one sig fig, I typed 187 as answer. The computer program told me I was "very close" but to check my rounding. What did I do wrong?