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CaHeK987 [17]
3 years ago
9

When comparing propionic acid (CH3CH2COOH) and sodium propionate (CH3CH2COONa), the one that is more soluble in water is When co

mparing propionic acid ( C H 3 C H 2 C O O H ) and sodium propionate ( C H 3 C H 2 C O O N a ), the one that is more soluble in water is blank.. When comparing cyclohexane (C6H12) and glucose (C6H12O6), the one that is more soluble in water is When comparing cyclohexane ( C 6 H 12 ) and glucose ( C 6 H 12 O 6 ), the one that is more soluble in water is blank.. When comparing hydrochloric acid (HCl) and ethyl chloride (CH3CH2Cl), the one that is more soluble in water is When comparing hydrochloric acid ( H C l ) and ethyl chloride ( C H 3 C H 2 C l ), the one that is more soluble in water is blank..
Chemistry
1 answer:
dedylja [7]3 years ago
6 0

Answer:

When comparing propionic acid (CH₃CH₂COOH) and sodium propionate (CH₃CH₂COONa), the one that is more soluble in water sodium propionate.

When comparing cyclohexane (C6H12) and glucose (C6H12O6), the one that is more soluble in water is glucose

When comparing hydrochloric acid (HCl) and ethyl chloride (CH3CH2Cl), the one that is more soluble in water is hydrochloric acid

Explanation:

When comparing propionic acid (CH₃CH₂COOH) and sodium propionate (CH₃CH₂COONa), the one that is more soluble in water sodium propionate. Because always a salt is more soluble in water than its acid (Solubility of propionic acid is 0,37g/mL and of sodium propionate is 1g/mL).

When comparing cyclohexane (C6H12) and glucose (C6H12O6), the one that is more soluble in water is glucose because cyclohexane is made from petroleum and its polarity is very low (cyclohexane is insoluble in water and solubility of glucose is 0,91g/mL).

When comparing hydrochloric acid (HCl) and ethyl chloride (CH3CH2Cl), the one that is more soluble in water is hydrochloric acid because an acid has a medium-high solubility in water but, again, an alkane derived from petroleum has very low solubility in water (hydrochloric acid has a solubility of 0,823g/mL and ethyl chloride is insoluble in water).

I hope it helps!

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Pavlova-9 [17]

Ferns use both sexual and asexual reproduction methods. In sexual reproduction, a haploid spore grows into a haploid gametophyte. ... The sporophyte produces spores, completing the life cycle. Asexual methods of reproduction include apogamy, poliferous frond tips, and rhizome spreading.

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4 years ago
Starting with lead (ii) carbonate describe how a solid sample of lead (ii) sulphate can be prepared
tatuchka [14]

The preparation of lead (ii) sulphate from lead (ii) carbonate occurs in two steps:

  1. insoluble lead carbonate is converted to soluble lead (ii) nitrate
  2. soluble lead (ii) nitrate is reacted with sulphuric acid to produce lead (ii) sulphate.

<h3>How can a  solid sample of lead (ii) sulphate be prepared from lead (ii) carbonate?</h3>

Lead (ii) carbonate and  lead (ii) sulphate are both insoluble salts of lead.

In order to prepare lead (ii) sulphate, a two step process is performed.

In the first step, Lead (ii) carbonate is reacted with dilute trioxonitrate (v) acid to produce lead (ii) nitrate.

  • PbCO₃ + 2HNO₃ → Pb(NO₃)₂ + CO₂ + H₂O

In the second step, dilute sulfuric acid is reacted with the lead (ii) nitrate to produce insoluble lead (ii) sulphate which is filtered and dried.

  • Pb(NO₃)₂ + H₂SO₄  →  PbSO₄ + 2HNO₃

In conclusion, lead (ii) sulphate is prepared in two steps.

Learn  more about lead (ii) sulphate at: brainly.com/question/188055

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4 0
2 years ago
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Read 2 more answers
When 1 mol of sodium nitrate, is dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed. What is the drop in
vesna_86 [32]

Answer:

 1.0  ° C

Explanation:

The molar mass for Sodium Nitrate NaNO₃ = (23+14+(16×3)) = 85

Number of moles of NaNO₃  = mass of NaNO₃ /molar mass of NaNO₃

⇒ 17/85 = 1.38 moles

Since 1 mole of NaNO₃  dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.

when 1.38 mole of NaNO₃  dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed..

Then; x kJ of 1.38 mole of NaNo₃ = 1.38 × 40 kJ  =55.2 kJ of heat absorbed.

Using the relation : Q = mcΔT  to determine the temperature drop ; we get:

55.2 = 17 × 4 (ΔT)

55.2  = 68 ΔT

ΔT= 0.8 ° C

ΔT ≅  1.0  ° C

Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is   1.0  ° C

7 0
3 years ago
10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0
4 years ago
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