PLEASE HELP!!! diluted titration calculation???!?
*** for the dilution *** M1V1 = M2V2 M2 = M1V1/V2 = 2.0 M x 5.0 mL / 250 mL = 0.040 M NaOH *** for the neutralization...*** idea here is moles H+ = moles OH- at neutralization... moles OH- can be found like this... 20 mL x (1 L / 1000 ml) = 0.020 L 0.020 L x (0.040 moles NaOH / L) = 0.00080 moles NaOH 0.00080 moles NaOH x (1 mole OH- / 1 mole NaOH) = 0.00080 moles OH- again, at neutralization, moles OH- = moles H+.. so moles H+ = 0.00080 moles H+ 0.00080 moles H+ x (1 mole HCl / 1 mole H+) = 0.00080 moles HCl finally, you have 0.020 M HCl... that means 0.020 moles HCl / L 0.00080 moles HCl x (1 L / 0.020 moles) = 0.040 L = 40 mL ie.. 40 mL of 0.020 M HCl was used. *********** alternately ********** when you get proficient in dimensional analysis.... 20 mL x (1L/1000 mL) x (0.040 moles NaOH / 1 L) x (1 mole OH- / 1 mole NaOH) x (1 mole H+ / 1 mole OH-) x (1 mole HCl / 1 mole H+) x (1L HCl / 0.020 moles HCl) x (1000 mL / L) = 40 mL HCl
Calc Help? Thank you!?
This is just iterated plug n chug I believe. The first thing you want to do, since (i) asks you to find it from [1,2]. First, run the equation when t =1, and it will tell you where the rock is it at time = 1. Then run it at t=2, and it will tell you where it is then. Take the difference of the value at t =2 and t =1 and you will find the distance. Velocity is dist/time, so you have found how many meters it covers in that second. Repeat this for all your different values. To find instantaneous velocity at t =1, use the pair [0,1] and do that same as above. I believe all of this should work. If not, sign up at NoteScoop.com and look through all of their stuff on math. They have worked out tests, quizzes, and tons of notes for math. They save me all the time.
Calc. Question. 10 points best ans. please help im desperate! thank you so much!?
Classic calculus question: lets do the dimensional analysis so you really understand L = length t= time L = L/t*L - L/t^2*t^2 To find the velocity function differentiate the position equation with respect to time V = 24 -1.6t L/t =L/t - L/t^2*t To find the acceleration function differentiate the velocity equation with respect to time A=-1.6 since acceleration is constant it is independent of time b) time to highest point V = 24 -1.6t = 0 24/1.6 = t = 15s c) s(t) =24t-0.8t^2 s(15) = 180m d) s(x) =24(r)-0.8*(r)^2 where x =180/2 = 90m we get a quadratic equation 0= 24(r)-0.8*(r)^2 - 90 solve for r e)Hint use the position equation set s(t) = 90m with the intitial velocity = 0 s = -0.8t^2 =90 solve for t add to the 15 second travel time up
I am poor in mathematics so how can I learn some fast calculation tips?
vedic(वैदिक) mathematics is a great tool for faster calculations. Late Shakuntala devi, also known as human computer, used to use it. It is a known fact that she could calculate faster than a super computer in America. You can also learn abacus used by Japenese and Chinese. Japanese are also known for calculation with the help of abacus with higher speed, even more than that of a calculator.
Modeling with first order equation!!!please help me!thank you very much!?
An important tool in archeological research is radiocarbon dating. This is a means of determining the age of certain wood and plant remains, hence of animal or human bones or artifacts found buried at the same levels. The procedure was developed by the American chemistWillard Libby (1908–1980) in the early 1950s and resulted in his winning the Nobel prize for chemistry in 1960. Radiocarbon dating is based on the fact that somewood or plant remains contain residual amounts of carbon-14, a radioactive isotope of carbon. This isotope is accumulated during the lifetime of the plant and begins to decay at its death. Since the half-life of carbon-14 is long (approximately 5730 years1), measurable amounts of carbon-14 remain after many thousands of years. Libby showed that if even a tiny fraction of the original amount of carbon-14 is still present, then by appropriate laboratory measurements the proportion of the original amount of carbon-14 that remains can be accurately determined. In other words, if Q(t) is the amount of carbon-14 at time t and Q0 is the original amount, then the ratio Q(t)/Q0 can be determined, at least if this quantity is not too small. Present measurement techniques permit the use of this method for time periods up to about 50,000 years, after which the amount of carbon-14 remaining is only about 0.00236 of the original amount. (a) Assuming that Q satisfies the differential equation Q_ = −rQ, determine the decay constant r for carbon-14. (b) Find an expression for Q(t) at any time t, if Q(0) = Q0. (c) Suppose that certain remains are discovered in which the current residual amount of carbon-14 is 20% of the original amount. Determine the age of these remains.
What are some tricks for fast calculation in Abacus?
For faster calculation in abacus, the number of fingers used to move the beads plays an important role. Most widely used fingers to calculate in abacus are index and middle finger in left hand and thumb and index finger in right hand. Calculation in usually done from the middle point if you are using Soroban. Interestingly, soroban consists of odd number of rods(vertical bars).The addition and subtraction calculations will be done from the middle point or middle rod. The key to calculate faster is to use index finger in right hand to make movements of heaven(upper) beads(both addition and subtraction), thumb for addition of earth(lower) beads and again index finger for subtraction. As for the left hand, for the movement of heaven beads, use middle finger for both addition and subtraction and for earth beads use index finger for both addition and subtraction.This technique is known as four finger technique.Most important thing to remember while using abacus is to make movements of all of your fingers simultaneously.EDIT:There is also six finger technique in which thumb index and middle fingers of both hands are used. middle finger for both addition and subtraction in upper beads, index fingers for subtraction in lower beads and thumb for addition in upper beads. The six fingers technique is faster than four fingers technique
How can I improve my calculation speed for SSC CGL exam?
Hello, this is a very important question and I am interested in reading answers given by others. All answers will be helpful in some or the other way. So thank you for asking this question.I was myself a banker and I have also cleared many competitive exams such as MPSC, RRB, SSC. I was also a Math faculty for competitive exams. In these years of experience of teaching and preparation, I have tried to raise my calculation speed using following remedies.By heart tables. It helps you to fast multiplication. In a second you can answer how much is 7 times 24.Practice oral additions and subtractions. First try single digit and then double digit. It helps a lot.Divisibility rules are like magic wand. Study them. They are easy to remember and easy to understand.When you know the tables and divisibility rules, then you can factorise in few seconds. That helps in finding HCF and LCM orally.It helps in addition and subtraction of fractions without pen and paper.It helps in reducing the ratios within no time.There are some ready percentage conversions. If you read the chart again and again, you will be fast in converting fraction to percentage and percentage to fraction orally.There are ratio tricks which are very useful for solving problems on ages orally. ‘These ratio tricks also help in mixture and alligations problems.Percentage conversion tricks are useful for profit and loss , interest type problems too.Byhearting squares upto 30 and cubes upto 10 is very useful.It helps you for finding square roots without pen and paper.I, myself like to keep discovering these ideas. I have my own channel which is absolutely free. Basically it covers quant for competitive exams and high school. In many topics, I have discussed all the above mentioned tricks. You can watch those videos to get an exact idea.Till now I have uploaded 58 videos and I am going to upload many more gradually. If you like the videos, you can subscribe ABSOLUTELY FREE.Just click below and start watching anytime. Practice is very important. Believe me, your brain gets developed accordingly and gets programmed for mental calculations automatically when you keep on practicing. Best wishes.Maths In Minutes with Priya
Pre-calc help: How do you convert 9 cis 300 degrees to rectangular coordinates?
9 is the hypotenuse of a right triangle and the diagonal of the rectangle you are looking for so y, the height, is 9sin(300)= -7.794 and x (width) is 9cos(300)=4.5
Applications of calc in the physical world? help???
A: x=50t-20t² B: x=80t² + 20t Note when t = 1.25hrs car A stops and goes back in the other direction! a) velocity is derivative of the distances given, and speed is the absolute value of velocity A: speed = | 50 - 40t | B: speed = | 160t + 20 | (This ones always positive so abs doesn't matter) at point O (t = 0) A = 10km/hr B = 20km/hr b) when does 50 - 40t = 160t + 20 200t = 30 t = 3/20 hours (or 9 minutes) c) Q = 50t-20t² = 80t² + 20t 100t² = 30t t = 0 (start) or t = 3/10 = 0.3hoursn (or 18minutes) Q = 50(3/10)-20(3/10)² = 13.2km d) If the third car is travelling at uniform speed then its velocity = that speed it starts at x = 2 v = s (its speed) x = st + c 2 = s(0) + c c = 2 x = st + 2 x = 13.2km when t = 3/10 hours (18 minutes) 13.2 = s(3/10) + 2 11.2 = 3s/10 s = 11.2 * 10/3 s = 112/3 km/hr (thats 27 and a third km/hr) x = 112t/3 + 2