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Sergio039 [100]
3 years ago
15

Write a balanced equation for the reaction of Na₂S and CdSO₄. Express your answer as a chemical equation. Identify all of the ph

ases in your answer. Enter no reaction if no precipitate is formed.
Chemistry
1 answer:
Luba_88 [7]3 years ago
3 0

Explanation:

A chemical reaction equation that contains same number of atoms on both reactant and product side is known as a balanced chemical reaction equation.

For example, Na_{2}S(aq) + CdSO_{4}(aq) \rightarrow Na_{2}SO_{4}(aq) + CdS(s)

Here, number of reactant molecules are as follows.

Na = 2

S = 2

Cd = 1

O = 4

Number of product molecules are as follows.

Na = 2

S = 2

Cd = 1

O = 4

As there are already equal number of atoms on both reactant and product side. Hence, the reaction equation is balanced.

Also, in this reaction both Na_{2}S and CdSO_{4} are present as aqueous solution. And, CdS is the precipitate that is formed as it is the insoluble solid that forms.

Na_{2}SO_{4} is present in aqueous state.

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Green plants absorb sunlight to power photosynthesis, the chemical synthesis of food from water and carbon dioxide. The compound
Kamila [148]

Answer:

698 \ THz

Explanation:

Data provided as per the question below:-

Wavelength = 430.nm

The computation of the frequency of the light is shown below:-

Frequency = Velocity of light ÷ Wavelength

The  Velocity of light = = {3.0 \times 10^8 \ m/s}

Wavelength = 430 nm = 4.30 \times 10^-^7 m

Frequency = \frac{3.0 \times 10^8 \ m/s}{4.30 \times 10^-^7}

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2 years ago
A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3
lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

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