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NorQuest College - Online Anytime
Канада
Добавлен 29 мар 2017
The NorQuest College - Online Anytime Channel offers curated and created videos for students in Academic Upgrading courses inside and outside of the college. These videos are both key and complementary learning resources for our Academic Upgrading courses.
I appreciate the video of how to analyze the poem. My teacher has been very vague and I am struggling in English concepts. Hopefully there’s more to comprehension, understanding and maybe writing a response to text!
Fact biomes
It is not available on play store.
Helpful time stamps: Part 1: 01:30 Initial colour of Bromocresol green 01:44 to 01:54 HCL added the first time 02:03 to 02:13 HCL added the second time 02:20 to 02:32 NaOH added the first time 02:39 to 02:47 NaOH added the second time Part 2: 04:02 10 drops KSCN potassium thiocyanate - test tube B 04:28 10 drops FeCl3 iron (III) cloride - test tube C 04:39 10 drops KCL potassium cloride - test tube D 04:54 10 drops NaOH sodium hydroxide - test tube E Part 3: 05:43 CoCl2 + HCL - first flask 05:52 CoCl2 + H2O (distilled) - second flask 06:04 water added to 1st flask 06:18 first flask heated 06:33 first flask cooled Part 4: 07:27 copper (III) sulfate added to erlenmyer flask 07:36 added ammonium hydroxide 07:51 added more ammonium hydroxide 07:59 added HCl
Thank you, that was a great video! Helped alot.
Arithmetic mistake! In c), my three subtotals (103776, 778320 and 1712304) are okay but somehow I didn't include the 778320 in my grand total. It should have come out to 2594400. Apologies - they call that "tripping at the finish line". :/
Thanks❤️
this saved my life bro
Thank you for this video. You are an amazing teacher. Your explanations are very clear and easy to understand.
Part 1 test data: 1:23 Mg(s)|Mg^2+(aq) 2:49 Cu(s)|Cu^2+(aq) 3:15 Al(s)|Al^3+(aq) 3:43 Sn(s)|Sn^2+(aq) 4:02 Fe(s)|Fe^3+(aq) 4:25 Ag(s)|Ag^+(aq) 4:43 Zn(s)|Zn^2+(aq) Part 2 test data: 5:22 connecting Aluminium and Tin half-cells. This is a reference for the experiment data so I don't have to scroll through the video to see the values again.
At 3:57 the anode was written as Ni(s) while the cathode was written as Mg(s) when at the beginning of the video at 2:26 it was the other way around. I'm assuming this is a mistake? It confused me for a minute there.
How is the answer to example 3 253 meters when the substitution is 27 - 11 / 2x1.2, which when put into a calculator and given brackets (to adhere to bedmas) equals 6.66666. . . repeating? I probably made a mistake but I can't see where.
thnx msr 💘💘
Okay what about sight what is it for you body to touch like for feeling
Okay then it to hard to now
he lived in his mom and dad's basement until he was 50 🤣
Thanks, this helped a lot! Could not figure it out on my own lol
And also thanks for ur great videos sir!
If it is heptanoic acid, shouldn't it have more hydrogen atoms than 6? 1:10
You're right. Heptanoic acid would be (pause for mental math) C6H13COOH. With five hydrogens this is probably *benzoic* acid, where you start with benzene (C6H6), remove one hydrogen and attach a carboxyl group in its place. Good catch - sorry for the mistake.
Thanks much
Thanks ☺️
Sound issues
Whoever this guy is, he may in fact be sparking the greatest academic comeback of my entire high school career.. God bless 🙏🙏
It's basement bob!
thank you
I just wanna point out that this video is the exact same as Physics 20 - Math skills for Physics. However, it is titled Relationships and Formulas. You might want to fix that.
Thanks for catching that!
For all the heads that r too cool for school, this guy thought me more than I learned in 2 months in one vid wrda like it up
Thank you! I now get it. I will add it at my poetry analysis assignment for 'Vigil I Kept on the field one Night' by W. Whitman,which an apostrophe itself and I had not paid attetion to it up to now!
This is a bit confusing because the book's solution says "This should be left as an exact value unless a decimal approximation is specifically requested or required". Also the books solution is: (125/400π)cm/s simplified to (5/16π)cm/s. Also it seems that (5/16π)cm/s would be roughly 0.9817cm/s but the video approximation is rounding to 0.1cm/s? @NorQuestCollegeOnlineAnytime can you help clear this up please.
im from philippines
Thanks for the video
Thank you for making such informative vedio
awesome
Watching from india ,we used to do such questions in our inter level chemistry examinations
I looked it up and couldn't find it
This was really helpful ❤
thanks Queen 😋
thanks queen 🙏 🙏🙏
Please note for Example 3, the answer is 40.0 m/s!
I need physics to be ultrasound technician, and have at least a 75% but I’m finding physics very challenging and hard to understand. I don’t know what I should do
Are you self-studying? If you're at a school, see what kind of tutorial support they have. Sitting down with a tutor periodically could help you fix any weak spots you're developing...
@@NorQuestCollegeOnlineAnytimeyes I’m self studying I’m going to try to get help
and the teachers at norquest are happy to help
useless asf
Ncert : Pea plant Tt gives tall plant (no mixing of traits) Ncert examplar: Rr gives pink flower
can't u explain why are they round, yellow, green smth?
Very helpful theres a video for each part of the diploma it helps alot seeing what we are asked to do and how we can approch the concept
This poem is written using several stanzas. Not only 4 stanzas
crazy this playlist doesnt have more watchs, super helpful man, thank you.
You're welcome! Hope it helped.
PKU is a homozygous recessive condition.suppose if one out of 10000 live birts are effected with PKU that what is the percentage of carriers in that population?
How come you did not solve for D on the last equation?
Nevermind Its the same method as solving for a.
I am loving these I calculated whole and thank you for making these video its really helping me
Note: on page 4 when we're asked to find the enthalpy changes for the reactions...we do want to do that, but what we're really going to need (to do Hess's law) is the MOLAR enthalpy changes, i.e. total enthalpy ÷ number of moles of reactant. Both of those numbers are between 100 and 500 kJ/mol, so if you're not landing in that range, check your math and get help if you need it! <3
How do we get the specific heat capacity for the reactants? I remember reading somewhere that since they're aq solutions, it's the same as water.
@@sansthedrummer That's right. Aqueous solutions are usually like 99% water by mass, so 4.19J/gC is usually close enough. :) n x delta-H is the energy in the chemical reaction, m x c x delta-T is the heating energy, and unless we have unusually bad experimental design, those amounts are equal.
@@JasonFahy thank you!! That was a sticking point for me.
Yes, BEDMAS still applies
does bedmas apply to this? when would we do the add/subracting first and when would we do the mul/dividing first?
yes
I watched the whole video looking for how to manipulate d=vft-1/2a(t^2) for t, but then you did it with vf=0. This does not help me a bit because I'm trying to figure out how to manipulate it when it still has two variables of t.
If you have a 't' term AND a 't^2' term, the the equation is quadratic and that makes the math a bit more painful - probably too painful for Physics 20. There are two scenarios. Scenario 1: d is zero. That helps a lot; we have 0 = v_ft - 1/2at^2, we can factor t out of it to obtain 0 = t(v_f - 1/2at) and that has two straightforward solutions, t=0 and t=2v_f/a. Scenario 2: d is NOT zero. Now the equation is a trinomial quadratic. If you're incredibly lucky it will be factorable, but generally we'd have to use the quadratic formula to find the value(s) of t. The formatting here is so limited, I don't think I can demonstrate the steps so I'll hope you know what I mean.
@@NorQuestCollegeOnlineAnytime I didn't realize at first, but most of the questions with d=vft-1/2a(t^2) that my teacher gives either have vf or vi = 0, or when that is not the case and I have to solve for t, I can use a different kinematic equation to find a new piece of information using my known variables. Then I can find an equation with the original variables plus the new one and solve it for t. I think I understand better now and thanks for taking the time to help.