All categories

3 years ago

Which of the following reactions is incorrectly labeled?

Chemistry

1 answer:

mars1129 [50]3 years ago

3 0

Answer:

The option a termed as precipitation reaction is incorrectly labelled. 

Explanation:

The chemical reactions are classified based on the reactants used and products formed in a reaction. They are decomposition reaction, single displacement reaction, double displacement reaction, acid-base neutralisation reaction, precipitation reaction, combustion reaction, redox reaction and organic reaction.

Among these, the given options are labelled as precipitation and combustion reaction. The one which is labelled as combustion reaction is correct because combustion reactions occur in the presence of oxygen only and the products of combustion reaction should include water, oxygen or carbon and heat.

The other option which is labelled as precipitation reaction is incorrect because precipitation reaction occurs when an ionic substance will come out of a solution due to heating it or stirring it making the solubility of the ionic substance in a solution zero such that it will come out as solid and form a layer at the bottom of the solution.

But in this case all the products are in aqueous state, there is absence of any ionic substance in solid state, so the option which is labelled as precipitation reaction is incorrectly labelled.

You might be interested in

A steel container with a movable piston contains 2.00 g of helium which was held at a constant temperature of 25 °C. Additional

shtirl [24]

Answer: D) 1.00 g

Explanation:

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

$V\propto n$

or,

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

where,

$V_1$ = initial volume of gas = 2.00 L

$V_2$ = final volume of gas = 3.00 L

$n_1$ = initial moles of gas = $\frac{\text {Given mass of helium}}{\text {molar mass of helium}}=\frac{2.00g}{4g/mol}=0.500mol$

$n_2$ = final moles of gas = ?

Now we put all the given values in this formula, we get

$\frac{2.00L}{3.00L}=\frac{0.500mol}{n_2}$

$n_2=0.75mole$

Mass of helium = $moles\times {\text {molar mass}}=0.75\times 4=3.00g$

Thus mass of helium added = (3.00-2.00) g = 1.00 g

4 0

3 years ago

What is the measurement for momentum?

sertanlavr [38]

The unit of momentum is the product of the units of mass and velocity.

8 0

2 years ago

A drop of vinegar ( a weak acid ) is placed on a sample of each of the materials below. Which will show the most active reaction

professor190 [17]

Answer: Option (B) is the correct answer.

Explanation:

Marble is also known as calcium carbonate and its chemical formula is $CaCO_{3}$ .

When it combines with vinegar then it results in the formation of calcium acetate which is a water soluble compound along with the release of carbon dioxide which occurs in the form of bubbles and water.

The reaction is as follows.

$CaCO_{3} + CH_{3}COOH \rightarrow Ca(CH_{3}COOH)_{2} + H_{2}O + CO_{2}$

Whereas vinegar being a weak acid will not react with glass, copper and steel.

8 0

3 years ago

Read 2 more answers

Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol

pishuonlain [190]

Answer: The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

Explanation:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

$0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g$

Calculating the freezing point:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.99°C/m

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC$

Hence, the freezing point of solution is -117.54°C

Calculating the boiling point:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_b$ = molal boiling point elevation constant = 1.20°C/m.g

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC$

Hence, the boiling point of solution is 80.48°C

3 0

3 years ago

The 488 nm laser is shine on a lithium metal whose work function is 2.9 eV will you be able to see any photoelectrons? if yes wh

nalin [4]

Answer:

There will not be any ejection of photoelectrons

Explanation:

Energy of the photon= hc/λ

Where;

h= Plank's constant

c= speed of light

λ= wavelength of the incident photon

E= 6.6×10^-34 × 3 ×10^8/488 × 10^-9

E= 4.1 ×10^-19 J

Work function of the metal (Wo)= 2.9 eV × 1.6 × 10^-19 = 4.64 × 10^-19 J

There can only be ejected photoelectrons when E>Wo but in this case, E<Wo hence there will not be any ejection of photoelectrons.

3 0

3 years ago