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Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Answer: Given Mass % of iron = 69.9 % Mass % of oxygen = 30.1 % Element Atomic mass Mass % Mass % / atomic mass Fe 55.85 69.9 69.9/55.85 = 1.25 O 16.00 30.1 30.1/16.00 = 1.88 Fe : O = 1.25 : 1.88 Converting...
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Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1
Solution: We have Mass percent of HNO3=69% Hence mass of HNO3 in the solution is 69g Mass of water=100-69=31g Number of moles of HNO3 =69/63=1.09moles Density=mass/volume Volume=mass/density =100/1.41=70.9ml Concentration (Molarity) = number of moles of solute/volume of the solution ...
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An oxide of an element was dissolved in water. The final solution turned red litmus blue. Is the element metal, non-metal or a metalloid?
The oxide of the element is basic as it turns red litmus blue .This means that the element is a metal (M).
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What are isoelectronic ions? Account for the decrease in size of following species oxide ion > Fluoride ion > sodium ion > magnesium ion.
The ions having same number of electrons are called isoelectronic ions. The above order is due to increasing order of effective nuclear charge.
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At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar.
Using the relationship of density (ρ) of the substance at temperature (T) We have ρ = Mp/RT or p = ρ RT/ M For the given data if M is the molar mass...
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Why is nitric acid added to Sodium Extract before adding Silvernitrate for testing halogens?
Nitric acid is added to decompose NaCN & Na2S. NaCN + HNO3 → NaNO3 + HCN Na2S + HNO3 → 2NaNO3 + H2S
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Write down the formula of (i) sodium oxide (ii) aluminium chloride (iii) sodium suphide (iv) magnesium hydroxide
(i) Sodium oxide → Na2O (ii) Aluminium chloride → AlCl3 (iii) Sodium suphide → Na2S (iv) Magnesium hydroxide → Mg(OH)2
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Write the chemical formula of the following: (a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate
(a) Magnesium chloride →MgCl2 (b) Calcium oxide →CaO (c) Copper nitrate →Cu(NO3)2 (d) Aluminium chloride →AlCl3 (e) Calcium carbonate →CaCO3
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Calculate the number of aluminium ions present in 0.051g of aluminium oxide. (Hint: The mass of an ion is the same
1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g i.e., 102g of Al2O3 = 6.022 × 1023 molecules of Al2O3 Then, 0.051 g of Al2O3...
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The amount of nitric acid required to oxidize 127 g of I2 to I2O5 will be? Assume that HNO3, gets converted to NO2.
The balanced chemical equation is given by I2 + 10HNO3 = I2O5 + 10NO2 + 5H2O Here 127 x 2 gm I2 is required to oxidise 63x10 gm of HNO3. So in 127 gm...
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