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Murljashka [212]
3 years ago
12

39. Which of the following substances are insoluble in water? Select all that apply.

Chemistry
2 answers:
Jet001 [13]3 years ago
7 0
A, B, and C are insoluble in water
Bond [772]3 years ago
5 0

Answer:

B) BaSO₄ and C) Mg₃(PO₄)₂

Explanation:

We can use the solubility rules to decide which substances are insoluble.

You have probably learned NAG SAG and "Cats Cradle Old People," Sally Said.

I've listed them below.

\begin{array}{lll}\textbf{Soluble} & \textbf{Insoluble}\\\textbf{N}\text{itrates} &\textbf{C} \text{arbonates }\\ \textbf{A}\text{cetates} & \textbf{C}\text{hromates}\\ \textbf{G}\text{roup 1} & \textbf{O}\text{ hydrOxides}\\ & \textbf{P} \text{hosphates}\\\textbf{S}\text{ulfates} &\textbf{S} \text{ulfites}\\ \textbf{A}\text{mmonium}& \textbf{S}\text{ulfides}\\\textbf{G}\text{roup 17} & \\\end{array}

These aids help you remember that:

The salts of Group 1 metals are soluble (Li₂CO₃ and KOH).

Sulfates are usually soluble, except for BaSO₄. Thus, BaSO₄ is insoluble.

Phosphates are insoluble — Mg₃(PO₄)₂

The two insoluble compounds are BaSO₄ and Mg₃(PO₄)₂.

 

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Water is flowing in a pipe with a mass flow rate of 100.0 lb/h (M water H,O=18.016). What is the (i) Molar flow rate of H20 (gmo
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Answer:

i) 0,7 molH20/s

ii)11,2 g O/s

iii)1,4 g H/s

Explanation:

i) To find the molar flow rate of water, we just convert the mass of water to moles of water using its molecular weight(g/mol) and changing to the proper units (lb to grames and hours to seconds):

100 \frac{lb}{h}*\frac{453,5g}{1 lb}*\frac{1molH20}{18,016g}*\frac{1h}{3600s}=0,7\frac{molH20}{s}

ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{1molO}{1molH20}*\frac{16gr}{1molO}*\frac{1h}{3600s}=11,2\frac{gO}{s}

iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):

100\frac{lb}{h} *\frac{453,5g}{1lb}*\frac{1 molH20}{18,016g}*\frac{2molH}{1molH20}*\frac{1gr}{1molH}*\frac{1h}{3600s}=1,4\frac{gH}{s}

3 0
2 years ago
The overall cell reaction occurring in an alkaline battery isZn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s) (e) In practice, vol
zubka84 [21]

b) Mass of MnO₂ = 5.981 g

    Mass of H₂O = 1.2384 g

c) Total Mass of Reactant consumed = 11.708 g

b) Given Reaction

            Zn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s)

  Mass of Zn = 4.50 g

  Moles of Zn = 0.0688 moles

   Now,

    Moles of Zn = moles of MnO₂ = moles of H₂O = moles of ZnO = moles of                   Mn(OH)₂

Hence ,

Moles of MnO₂ = 0.0688 moles

Mass of MnO₂ = 0.0688 × 86.9368 g

                        = 5.981 g

Similarly,

    Moles of H₂O = 0.0688 moles

     Mass of H₂O = 0.0688 × 18 g

                           = 1.2384 g

c) now ,

    Moles of  ZnO = 0.0688 moles

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Hence,

    Total mass of reactant consumed = 11.708 g

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G = - nFEcell

 

and no. of moles of reactant  is greater than that of number of moles of reactant consumed .

       Hence voltaic cell of given Capacity are heavier than that of mass of reactant consumed .

 Thus from above conclusion we can say that , Mass of the reactant consumed is 11.708 g.

Learn more about Galvanic Cell here : brainly.com/question/19340007

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