Some fruits and vegetables are preserved by pickling them. Nandini got confused

<span>To determine the balanced chemical equation, it is necessary to identify the symbols and charges for each element and then balance the equation. This results in: Al(OH)3(aq) + 3HBr(aq) ---> AlBr3(aq) + 3H2O(l) The net ionic equation involves summing the charges and breaking them down into fundamental parts. Al3+ + 3OH- + 3H+ + 3Br- --> Al3+ + 3Br + H3O+ After eliminating aluminum and bromine, we have: 3OH-(aq) + 3H+(aq) --> 3H2O(l)</span>

Explanation:

Elements provided:

  F, Sr, P, Ca, O, Br, Rb, Sb, Li, S

Elements sharing similar reactivity belong to the same group in the periodic table, indicating that those in the same column exhibit comparable reactivity. Here are the identified groupings:

  Li and Rb are alkali metals in group 1

  Ca and Sr are alkaline earth metals in group 2

  F and Br are halogens in group 7

  O and S belong to group 6

 P and Sb are classified in group 5 of the periodic table

Thus, these classifications illustrate elements with the same chemical characteristics.

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g × = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g × = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

Response:

The specific heat of the alloy

Clarification:

Weight of the alloy = 25 gm

Initial temperature = 100°c = 373 K

Weight of the water = 90 gm

Initial temperature of water = 25.32 °c = 298.32 K

Final temperature = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

[[ - ] = ( - )

25 × × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

This gives us the specific heat of the alloy.

Answer:

710.33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 utilized for combusting octane

= 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2