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Zanzabum
3 years ago
15

A sample of a substance that has a density of 0.824 g/mL has a mass of 0.451g. Calculate the volume of the sample.

Chemistry
1 answer:
Allushta [10]3 years ago
8 0
Density can be calculated using the following rule:
density=mass/volume
therefore,
volume=mass/density

we have mass=0.451g and density=0.824g/ml
substituting in the above equation, we can calculate the volume as follows:
volume = 0.451/0.824 = 0.547 ml
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For the diprotic weak acid h2a, ka1 = 3.2 × 10-6 and ka2 = 6.1 × 10-9. what is the ph of a 0.0650 m solution of h2a? what are th
Stolb23 [73]
Given:

Diprotic weak acid H2A:
 
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9. 
Concentration = 0.0650 m 

Balanced chemical equation:

H2A ===> 2H+  + A2- 
0.0650       0        0
-x                2x       x
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0.065 - x     2x      x

ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)

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3 years ago
Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,
torisob [31]

Answer : The full balanced ionic equation will be,

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, K^+,NO_3^-

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)

The ionic equation in separated aqueous solution will be,

2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)

In this equation, K^+\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)

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