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3 years ago

Explain how you can obtain hydrated sodium sulphate crystals from an aqueous solution of sodium sulphate

Chemistry

2 answers:

Mice21 [21]3 years ago

6 0

Answer:

If you left your aqueous sodium sulfite solution open on the side, atmospheric oxygen would cause the chemical change, the water would hydrate the new compound and the slow evaporation of water would give you lovely large crystals of Na2SO4 without you having to do anything!

lana66690 [7]3 years ago

6 0

Answer:

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Which element in the list is the most reactive? C Ba S Cs

Tems11 [23]

Answer:

Explanation:

This element is the farthest right, and lowest meaning that the reactivity of it is very high.

1. valence electrons and electronegativity

2. Both reactivity and electronegativity decrease down a group.

3. Electronegativity decreases down a group, and reactivity increases down a group.

4. The number of valence electrons increases from left to right.

5. Cs

{quizlet: captncrun}

4 0

2 years ago

Consider the balanced equation. 2HCl + Mg MgCl2 + H2 If 40.0 g of HCl react with an excess of magnesium metal, what is the theor

Serga [27]

Answer:

Theoretical yield of hydrogen is 1.11 g

Explanation:

Balanced equation, $Mg+2HCl\rightarrow MgCl_{2}+H_{2}$

As Mg remain present in excess therefore HCl is the limiting reagent.

According to balanced equation, 2 moles of HCl produce 1 mol of $H_{2}$ .

Molar mass of HCl = 36.46 g/mol

So, 40.0 g of HCl = $\frac{40.0}{36.46}moles$ of HCl = 1.10 moles of HCl

Hence, theoretically, number of moles of $H_{2}$ are produced from 1.10 moles of HCl = $(\frac{1}{2}\times 1.10)moles=0.550moles$

Molar mass of $H_{2}$ = 2.016 g/mol

So, theoretical yield of $H_{2}$ = $(0.550\times 2.016)g=1.11g$

7 0

4 years ago

Read 2 more answers

Which is an example of a physical change?

levacccp [35]

Answer:

salt dissolving

Explanation:

Hope this helps!

8 0

3 years ago

Read 2 more answers

a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v

Rashid [163]

Concentration of NaOH = 0.148 molar, M

Concentration of H3PO4 = 0.172 molar, M

Concentration x Volume will give the number of moles of solute in that volume. C*V = moles

Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:

NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:

(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M

====

The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:

(0.00172 moles H3PO4)/(0.010 liters H3PO4)

Concentration of H3PO4 = 0.172 molar, M

5 0

1 year ago

Did I do this science worksheet correctly? Asking for a friend

grin007 [14]

Yes everything looks good to me

8 0

3 years ago

Explain how you can obtain hydrated sodium sulphate crystals from an aqueous solution of sodium sulphate​

Explain how you can obtain hydrated sodium sulphate crystals from an aqueous solution of sodium sulphate