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Sergio039 [100]
4 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]4 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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How many valence electrons does K+ have
Lera25 [3.4K]

Answer:

About one valence electron

Explanation:

Obviously, removing that electron gives us [Ar] (same configuration as K1+), which is a noble gas and has 8 electrons. Valence electrons are generally regarded as being 'the outermost electrons' for a given atom. Therefore, with neutral potassium, there is one valence electron

5 0
3 years ago
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Which statements about thermal energy are true? Choose all answers that are correct. A. Temperature is a measure of the average
Alekssandra [29.7K]
The answers that are correct are a, b, and d
3 0
3 years ago
A certain substance melts at a temperature of . But if a sample of is prepared with of urea dissolved in it, the sample is found
pshichka [43]

Answer:

2.2 °C/m

Explanation:

It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:

" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "

So we use the formula for <em>freezing point depression</em>:

  • ΔTf = Kf * m

In this case, ΔTf = 13.2 - 9.9 = 3.3°C

m is the molality (moles solute/kg solvent)

  • 350 g X ⇒ 350/1000 = 0.35 kg X
  • 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea

Molality = 0.53 / 0.35 = 1.51 m

So now we have all the required data to <u>solve for Kf</u>:

  • ΔTf = Kf * m
  • 3.3 °C = Kf * 1.51 m
  • Kf = 2.2 °C/m
5 0
3 years ago
PLEASE HELP!!! I will mark brainliest I am really struggling with these problems
Ket [755]

Answer:

21883.75 Joules are required to melt the ice!

5 0
3 years ago
How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr
andreyandreev [35.5K]

Answer:

percentage yield = 67%

Explanation:

Mass of Cu(NO₃)₂  = 15.25 g

Mass of NaOH   = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH   →  Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.      NaOH             :      Cu(OH)₂

                               2                   :          1

                               0.32              :           1/2×0.32 = 0.16 mol

                            Cu(NO₃)₂         :           Cu(OH)₂

                                  1                  :               1

                             0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

5 0
3 years ago
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