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Margaret [11]
3 years ago
13

Mohr's salt is a pale green crystalline solid which is soluble in water. It is a 'double sulfate' which contains two cations, on

e of which is Fe2+.The identity of the second cation was determined by heating solid mohr's salt with solid sodium hydroxide and a colourless gas was evolved. The gas readily dissolved in watdr giving an alkaline solution. A grey green solid residue was also formed which wasd insoluble in water. Whate are the identities of the gas and solid residue? A. Gas: H2 Residue: FeSO4 B.NH3 Na2SO4 C. NH3 Fe(OH)2 D. SO2 Fe(OH)2
Chemistry
1 answer:
Veronika [31]3 years ago
3 0
The gas is NH₃.
H₂ doesn't dissolve readily in water, SO₂ gives an acidic solution in water.

The solid residue is Fe(OH)₂.
FeSO₄ and Na₂SO₄ are soluble in water.

The answer is C.
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Chemical reaction when chromium metal is immersed in an aqueous solution of cobalt(II) chloride.
Marrrta [24]

2Cr + 3CoCl_2→ 2CrCl_3 + 3Co

Explanation:

  • The products formed are chromic chloride and cobalt.

        Chromium + Cobaltous Chloride = Chromic Chloride + Cobalt

  • Type  of reaction is Single Displacement (Substitution) which is there is a displacement of one atom.

Reactants used in the reaction are -

  • Chromium(Cr)
  • Cobaltous Chloride (CoCl_2)

Products formed in the reaction are -

  • Chromic Chloride(CrCl_3)
  • Cobalt (Co)

Hence, the chemical reaction is as follows -

  • Cr + CoCl_2 →CrCl_3 + Co

For balancing the above chemical equation we need to add a coefficient of 2 in front of chromium and of 3 in front of cobalt(II)chloride on right-hand-side while of 2 in front of chromium chloride and of 3 in front of carbon monoxide on left-hand-side of the equation.

Hence, the balanced equation is -

2Cr + 3CoCl_2→ 2CrCl_3 + 3Co

3 0
3 years ago
How many carbon atoms are in 15.0 grams of sucrose (C12H22O11)
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Answer: 15.3 carbon atoms
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If water was used to rinse the burette used to dispense the hydrochloric acid solution, how would this affect
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Answer:

The concentration of hydrochloric acid would be estimated to be less.

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7 0
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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

4 0
3 years ago
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