Top answers

Chemistry
GCSE

Compare the structure between graphite and diamond

KeypointsEach diamond carbon is covalently bonded to 4 other carbonsEach graphite carbon is covalently bonded to 3 other carbonsDiamond is very hard due to its tertrahedral structureGraphite is made up of...

Answered by Hamza A. Chemistry tutor
1385 Views

CO(g) + 2H2(g) ⇌ CH3OH(g). A pressure of 100 atmospheres is used instead of atmospheric pressure. The higher pressure gives a greater yield of methanol and an increased rate of reaction. Explain why.

4 marksExplanation: Increasing the pressure (gases) of reactions increases the frequency of collisions, therefore increasing the rate of reaction.If the ...

Answered by George O. Chemistry tutor
6180 Views

propene and steam react to form propan-2-ol. If the yield is 75% what mass of propene is required to produce 410g of propane-2-ol. [molar masses/gmol-1 propene=42, propan-2-ol =60]

remember the equation moles = mass/molar mass. We have the final grams of product and also the molar mass so we can calculate the number of moles we got from this. [410/60=6.83moles]

Answered by Jasmine W. Chemistry tutor
4824 Views

In Stage 2 of a reaction, 40 kg of titanium chloride was added to 20 kg of sodium. The equation for the reaction is: TiCl4 + 4Na -> Ti +4NaCl Relative atomic masses (Ar): Na = 23 Cl = 35.5 Ti = 48 Explain why titanium chloride is the limiting reactant.

To calculate which of titanium chloride or sodium is limiting, convert both into mol so they are directly comparable. You cannot compare in masses because in equations there is a conservation of moles and...

Answered by Simran L. Chemistry tutor
13697 Views

If we have 10 grams of Helium at a concentration of 10 mol dm-3, what volume of helium do we get.

First we need to identify that 2 equations will be in use here. 1st - moles = mass/Mr and 2nd - volume = moles/concentration This is because we need to find the volume and the only other information we ha...

Answered by Hamaad H. Chemistry tutor
1486 Views

We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences