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ASHA 777 [7]
3 years ago
12

Provide one example where a neutralization reaction is undesired/harmful. State the reactants and explain why it is harmful for

this reaction to take place.
Chemistry
1 answer:
Crazy boy [7]3 years ago
8 0

Answer:

You're in my science class wth

Explanation:

For instance farmers are using basic fertilizers to neutralize an acid soil –which is harmful for the plants- while basic solutions are added in the chimneys of factories to neutralize the acid smoke that they produce –avoiding the catastrophic effects of the acid rains. Acid released in the environment from industry creates acid rain which has many ecological effects. Acid rain makes waters acidic, and causes them to absorb the aluminium that makes its way from soil into lakes and streams. This combination makes waters toxic to crayfish, clams, fish, and other aquatic animals.

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When water and alcohol are mixed, the final volume is less than the total of volume of alcohol plus water added due to .......​
tigry1 [53]

Answer:

molecules take up more space

6 0
3 years ago
Predict the product formed when the compound shown below undergoes a reaction with mcpba in ch2cl2. mcpba is meta-chloroperoxybe
Aleks04 [339]
Alkenes on reaction with meta-chloroperoxybenzoic acid (MCPBA ) produces epoxides. When styrene is reacted with mCPBA it gives 2-phenyloxirane as shown below,

6 0
3 years ago
Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of glucose.
Rainbow [258]

Answer:

-0.1767°C (Option A)

Explanation:

Let's apply the colligative property of freezing point depression.

ΔT = Kf . m. i

i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1

m = molality (mol of solute / 1kg of solvent)

We have this data → 0.095 m

Kf is the freezing-point-depression constantm 1.86 °C/m, for water

ΔT = T° frezzing pure solvent - T° freezing solution

(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1

T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C

4 0
3 years ago
the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel
shtirl [24]

<u>Answer:</u> The entropy change of the process is 2.0\times 10^2J/K

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{f}}{T}

where,  

\Delta S = Entropy change

n = moles of acetone = 6.3 moles

\Delta H_{f} = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol    (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = -94.7^oC=[273-94.7]=178.3K

Putting values in above equation, we get:

\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K

Hence, the entropy change of the process is 2.0\times 10^2J/K

4 0
4 years ago
If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction
Tasya [4]

Answer:

<u></u>

  • <u>0.456M</u>

Explanation:

<u>1. Balanced molecular equation</u>

     2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O

<u>2. Mole ratio</u>

     \dfrac{2molHNO_3}{1molBa(OH)_2}

<u>3. Moles of HNO₃</u>

  • Number of moles = Molarity × Volume in liters
  • n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

<u>4. Moles Ba(OH)₂</u>

  • n = 0.700M × 0.0310 liter = 0.0217 mol

<u>5. Limiting reactant</u>

Actual ratio:

   \dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

<u>6. Final molarity of Ba(OH)₂</u>

  • Molarity = number of moles / volume in liters
  • Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
5 0
3 years ago
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