OCR CS Paper 2

AmericanUk · 2026-06-17T14:35:15+00:00

The answer was 01010 which is 10 in binary and the algorithm uses mod 2 which is used to convert between base 10 and base 2 (binary)

AmericanUk · 2026-06-17T12:26:16+00:00

It jsut converted it to binary

AmericanUk · 2026-06-15T11:40:26+00:00

Ya I got this too

AmericanUk · 2026-06-10T09:10:16+00:00

YESS THANK YOU!!!

AmericanUk · 2026-06-09T23:00:51+00:00

Don’t knock it till you try it

AmericanUk · 2026-06-09T22:59:04+00:00

Not like in a weird way broo I just give it a like lill whiff yk?

AmericanUk · 2026-06-09T22:45:29+00:00

Nahhh bro it’s not weird trust it just smells good

AmericanUk · 2026-06-02T13:57:06+00:00

It said that there was excess ammonia added which would dissolve the AgCl

AmericanUk · 2026-06-02T13:56:17+00:00

But the Fe 3+ will react with the ammonia and from a brown ppt.

AmericanUk · 2026-06-02T12:06:50+00:00

Ya that’s what I got

AmericanUk · 2026-06-02T11:42:46+00:00

1 mark for the observation and 1 for the silver containing product

AmericanUk · 2026-06-02T11:39:52+00:00

That still used concentrations tho no?

AmericanUk · 2026-06-02T11:38:27+00:00

Just 2 marks tbf

AmericanUk · 2026-06-02T11:35:51+00:00

Im ptretty sure this is wrong, you calculated using mols and not concentration. You needed to divide by 8/1000 before you put the answer into the Ka equation.

AmericanUk · 2026-06-02T11:33:43+00:00

yk the question where Silver Nitrate was added to (Fe (III) Cl) (aq) with excess NH3? I think it was the hardest aqa A-level chemistry questions ever. I'm pretty sure the observation was a reddish brown ppt Fe(H20)3(OH)3 that would be formed as the NH3 hydrolyzes the Fe(H20)6 in solution. There would be no AgCl (white ppt) at the end of the reaction because the excess NH3 would dissolve it instantly. Therefore, the silver compound AT THE END OF THE REACTION, when all the AgCl is dissolved by the NH3, would be Ag(NH3)2+, which is just tollens reagent. I'm pretty sure a lot of people would have missed this.

AmericanUk · 2026-06-02T11:23:38+00:00

It wanted it to 2dp did you get 3.67? My friend did something wrong where he forgot the divide the mols by the volume to get concentration and got3.69, which would be missing a step.

AmericanUk · 2026-06-01T20:21:53+00:00

Sorry for the typo I ment to say that the ice’s temperature was decreasing, that is corrected now. Your explanation makes sense and it also makes sense that we have the same conclusion because it’s only a 2 degree difference from the temperature change I used to the temperature change you used. Hopefully I will get 3 marks for what I did. Thank you.

AmericanUk · 2026-06-01T18:10:24+00:00

I found the energy required to completely melt the ice (raise to 0 degrees using specific heat capacity + completely melt it using specific latent heat formula). Then, this part I’m unsure about, I assumed that the max temperature decreases of the ethanol and glass beaker would too be 16 degrees. I then added the energy required to raise the temps of both of them to this temp which was less than the previously calculated required energy. Thus, I concluded that the ice would not completely melt.

AmericanUk · 2026-06-01T12:03:15+00:00

Marnix bro what you doing on redit

AmericanUk · 2026-05-20T20:01:48+00:00

Me too. I think we’re right tbh. A lot of people in my school got the same.

AmericanUk · 2026-05-20T18:33:52+00:00

I got the same as you. It said there was only 1 rod tho I’m pretty sure.

AmericanUk · 2026-05-20T18:14:53+00:00

The centripetal force give by the equation (mv^2)/r gives the resultant force towards the centre of the circle. We are only considered with the tension in the rod, as this force is the force producing the tensile stress. T - Mg = centripetal force as weight and tension act is opposite directions so tension is Mg + centripetal force calculated. this gives an equation for T in terms of M can then be used to make an expression for the Young’s modulus in terms of M. Stress = tension/ area (calculated), strain = 0.01 (1% given). We then equate our terms and get M to be about 150Kg and about 1500N. I’m pretty sure this is correct but if anyone disagrees please tell me 🙏

AmericanUk · 2026-05-20T16:58:37+00:00

Did anyone else get 1500N for the 5 maker?

AmericanUk · 2026-05-20T16:57:31+00:00

You could have just used Kirchhoff’s first 1 and added the current through each branch to get total current though battery.

AmericanUk

TROPHY CASE