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scZoUnD [109]
3 years ago
13

What are the answers to the blanks to make this equation balanced

Chemistry
1 answer:
Eddi Din [679]3 years ago
5 0

1Al+6NaOH==>2Na_{3}AlO_{3}+3H_{2}


Hope this helps!

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The sp of pbbr2 is 6. 60×10−6. what is the molar solubility of pbbr2 in pure water?
Inessa05 [86]

Ksp of PbBr₂ is 6.60 × 10⁻⁶. The molar solubility of PbBr₂ in pure water is 0.0118M.

Ksp or Solubility Product Constant is an equilibrium constant for the dissociation in an aqueous solution.

Molar solubility (S) is the concentration of the dissolved substance in a solution that is saturated.

Let the molar solubility be S upon dissociation.

PbBr₂ or Lead Bromide dissociates in pure water as follows:

                          PbBr₂ ----------> Pb⁺² + Br⁻

                                                     S      2S

Ksp = [Pb⁺²] [ Br⁻]

Ksp = (S) (2S)²

Ksp = 4S³

6.60 × 10⁻⁶ = 4S³

S = 0.0118M

Hence, the Molar solubility S is 0.0118M.

Learn more about Molar solubility here, brainly.com/question/16243859

#SPJ4

6 0
2 years ago
Binigyan ang isang liham ng pasasalamat at gagawin ko rin siya ng kabutihan sa ibang pagkakataon​
anzhelika [568]

Answer:

what language do you speak?

8 0
2 years ago
Find the empirical formula of the following compounds:
Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus \frac{mol phosphorus}{ 30.97 g phosphorus}= 0.0293 mol

Moles bromine 6.99 g bromine\frac{mol bromine}{79.90 g bromine}=0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

8 0
1 year ago
Each step in the following process has a yield of 70.0%. CH4+4Cl2⟶CCl4+4HCl CCl4+2HF⟶CCl2F2+2HCl The CCl4 formed in the first st
Nadusha1986 [10]

Answer:

n(HCl)=1.96 mol

Explanation:

CH4+4Cl2⟶CCl4+4HCl

CCl4+2HF⟶CCl2F2+2HCl

With ideal yields we will end up with 4 moles of HCl.

With 70% yields on every stage

n(HCl)=0.7*0.7*4=1.96 mol

8 0
3 years ago
What method is used to removed the unreacted copper (II) carbonate?​
Evgen [1.6K]

Answer:

Filtration

Explanation:

Metal carbonate is insoluble, it is possible to filter off the unreacted substances leaving the desired salt solution

8 0
3 years ago
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