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3 years ago

When heat energy is removed, two changes are observed:

Chemistry

2 answers:

Valentin [98]3 years ago

8 0

Answer : The correct option is, A is condensation and B is freezing.

Explanation :

Melting or fusion : It is a type of process in which the phase changes from solid state to liquid state at constant temperature.

Freezing : It is a type of process in which the phase changes from liquid state to solid state at constant temperature.

Evaporation : It is a type of process in which the phase changes from liquid state to gaseous state at constant temperature.

Condensation : It is a type of process in which the phase changes from gaseous state to liquid state at constant temperature.

Sublimation : It is a type of process in which the phase changes from solid state to gaseous state without passing through the liquid state at constant temperature.

Deposition : It is a type of process in which the phase changes from gaseous state to solid state without passing through the liquid state at constant temperature.

The given observations are:

Change A: Gas changes phases to a liquid. → Condensation

Change B: Liquid changes phases to a solid. → Freezing

Hence, the correct option is, A is condensation and B is freezing.

harkovskaia [24]3 years ago

3 0

Answer: Option (A) is the correct answer.

Explanation:

Condensation is defined as the process in which vapors or gaseous phase changes into liquid phase.

Freezing is the process in which liquid phase changes into solid phase.

Evaporation is the process in which liquid phase changes into vapors or gaseous phase.

Melting is the process in which solid phase changes into liquid phase.

Therefore, we can conclude that names for given phase changes A and B is that A is condensation and B is freezing.

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1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate

Whitepunk [10]

a. 1,4332 g

b. 7.54~g

<h3>Further explanation</h3>

Given

Reaction

MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

$\tt 20~cm^3=20\times 10^{-3}~dm^3\\\\mol=M\times V\\\\mol=2.5~mol/dm^3\times 20\times 10^{-3}DM^3=0.05$

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

$\tt mol=M\times V\\\\mol=\dfrac{2.5~g/dm^3}{95,211 g/mol}=0.0263~mol/dm^3$

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g

6 0

2 years ago

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

<u>Answer:</u> The percent yield of the reaction is 32.34 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For benzoic acid:</u>

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

6 0

3 years ago

The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its

Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

$t_{\frac{1}{2}} =$ $\frac{ln 2}{k}$

$t_{\frac{1}{2}} = \frac{0.693}{k}$

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

$ln(0.35) = -(1.325 *10^{-4})(t)$

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0

3 years ago

What is the enthalpy of reaction?

erma4kov [3.2K]

Answer:

Enthalpy is the amount of heat released or absorb. The total enthalpy can be found by subtracting the products' enthalpy by the reactants' enthalpy

4 0

3 years ago

Read 2 more answers

ODIO POD

miss Akunina [59]

Answer:

$\boxed{\text{40 mol Al}}$

Explanation:

Al₂O₃ ⟶ 2Al + 3O₂

n/mol: 20

$\text{Moles of Al} = \text{20 mol Al$_{2}$O$_{3}$}\times \dfrac{\text{2 mol Al}}{\text{1 mol Al$_{2}$O$_{3}$}}= \textbf{40 mol Al}\\\text{You can produce }\boxed{\textbf{40 mol Al}}$

8 0

3 years ago