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Question 1 Report
Aluminum hydroxide is used in the dyeing industry as a?
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Question 8 Report
A few drop of conc HCL are added to about \(10\text{cm}^{3}\) of a solution of PH 3.4. The PH of the resulting mixture is
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Question 12 Report
The number of molecules of Carbon(iv)Oxide produced when 10.0g of \( \mathrm{CaCO_3} \) is treated with \(0.2\mathrm{dm^3}\) of 1 Mole of HCL in the equation
\[ \mathrm{CaCO_3 + 2HCL \rightarrow CaCl_2 + H_2O + CO_2} \], is ?
Question 13 Report
Which of the gas laws does this graph illustrate?
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Question 17 Report
Using the combined gas law formula,
Given that P1 = 0.825, V1 = 2.76 L, V2 = 1.38 L, P2 = ?
And T1 = T2; we would have P2 = [ P1 X V1 ] / V2
: [0.825 X 2.76] / 1.38
= 1.650atm
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Question 19 Report
A gaseous metallic chloride MCl consists of 20.22% of M by mass. The formula of the chloride is?
[ M = 27, Cl = 35.5]
| M | Cl | |
| % composition | 20.22 | 79.78 |
| Atomic mass | 27 | 35.5 |
| Mole ratio | 20.22 | 79.78 |
| 27 | 35.5 | |
| 0.75 | 2.25 | |
| Divided | 0.75 | 0.75 |
| 1 | 3 |
The formula of the Chloride = MCl3
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Question 23 Report
4OH ⇒ 2H2 O + O2 + 4e-
From the equation:
1 mole of Oxygen requires 4e
i.e 1 mole of Oxygen requires 4F,
where 1F = 96,500 C
[ 4 X 96,500 X 0.125 ] ÷ 1
= 48,250 C
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Answer Details
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Question 27 Report
8g of \( \mathrm{CH_4} \) occupies 11.2 at S.T.P. What volume would 22g of \( \mathrm{CH_3CH_2CH_3} \) occupy under the same condition?
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Question 29 Report
How many valence electrons are contained in the element \( {}_{15}^{31}\mathrm{P} \)?
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Question 33 Report
The volume occupied by 1.58g of a gas at S.T.P is 500cm\(^{3}\). What is the relative molecular mass of the gas? [ G.M.V at S.T.P = 22.4dm\(^{3}\) ]
Question 34 Report
The electronic configuration of an element is \(1S^2\) \(2S^2\) \(2P^6\) \(3S^2\) \(3P^3\). How many unpaired electrons are there in the element?
Once you figure out the electron configuration, you fill up the corresponding orbitals with electrons, any left with one is considered unpaired. Since 1s can only hold 2 electrons, and P has 15, that's obviously filled and has no unpaired electrons. The same is for 2s which holds 2, 2p which holds 6, 3s which holds 2.However 3p can hold 6 electrons and in order for that to be filled up you would need to have an element of 18 electrons. So you fill up as much as you can in 3p by first adding 1 electron to each energy level. 3p has 3 energy levels and there are only 3 electrons left to distribute, so each of those energy levels only gets 1, because you have to fill them all with one before you can start adding a second.So since you are only able to fill one electron in each of the three energy levels of the 3p orbital, that leaves the orbital open for 1 more electron in each of its energy levels. So there are 3 unpaired electrons in P.
Question 35 Report
The general formula for the Alkanals is?
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Question 40 Report
32g of anhydrous copper(ii)tetraoxosulphate(vi) dissolved in \(1\mathrm{dm}^3\) of water generated 13.0kj of heat. The heat solution is?
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