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4 years ago

14

A positively charged ion is called a cation which means there are more protons in the atom compared to the number of protons in

the neutral atom. True False

1 answer:

Lelechka [254]4 years ago

7 0

False False False False False False Ф

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Jane described two substances which participate in photosynthesis.

Anika [276]

Answer: -

Substance A is carbon dioxide and Substance B is oxygen.

Explanation: -

Photosynthesis reaction involves the process of conversion of light energy of the sun into chemical energy by the green chlorophyll found in plants leaves.

The chemical equation for the process is

6 CO₂ +6 H₂O → C₆H₁₂O₆ + 6O₂

As we can see Carbon dioxide CO₂ reacts with water H₂O to form glucose. Oxygen is the gaseous product formed.

Thus Substance A is carbon dioxide and Substance B is oxygen.

8 0

4 years ago

The following equations are half reactions and reduction potentials. Ag+ (aq) + e– Ag(s) has a reduction potential of +0.80 V. C

Lady bird [3.3K]

Answer : The correct option is, (A) silver ion gains electrons more easily and is a stronger oxidizing agent than a chromium(III) ion.

Explanation :

The given half reaction are :

1st half reaction : $Ag^+(aq)+e^-\rightarrow Ag(s)$

The reduction potential of this reaction = +0.80 V

2nd half reaction : $Cr^{3}+(aq)+3e^-\rightarrow Cr(s)$

The reduction potential of this reaction = -0.74 V

From the reduction potentials, we conclude that the reaction which have positive reduction potential, they will gain electrons more easily and reduced itself and act as a stronger oxidizing agent.

Or we can say that the reaction which have negative reduction potential, they will lose electrons more easily and oxidized itself and act as a stronger reducing agent.

The reduction potential of 1st half reaction is positive and 2nd half reaction is negative.

Therefore, the silver ion gains electrons more easily and is a stronger oxidizing agent than a chromium(III) ion.

5 0

4 years ago

Read 2 more answers

The normal boiling point of ethanol (C2H5OH) is 78.3 oC and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the chan

Sloan [31]

Answer:

The change in entropy in the system is -231.5 J/K

Explanation:

Step 1: Data given

The normal boiling point of ethanol (C2H5OH) is 78.3 °C = 351.45 K

The molar enthalpy of vaporization of ethanol is 38.56 kJ/mol = 38560 J/mol

Mass of ethanol = 97.2 grams

Pressure = 1 atm

Step 2: Calculate the entropy change of vaporization

The entropy change of vaporization = molar enthalpy of vaporization of ethanol / temperature

The entropy change of vaporization = 38560 J/mol / 351.45 K

The entropy change of vaporization = -109.72 J/k*mol

Step 3: Calculate moles of ethanol

Moles ethanol = mass / molar mass ethanol

Moles ethanol = 97.2 grams / 46.07 g/mol

Moles ethanol = 2.11 moles

Step 4: Calculate change in entropy

For 1 mol = -109.72 J/K*mol

For 2.11 moles = -231.5 J/K

The change in entropy in the system is -231.5 J/K

7 0

3 years ago

|A titanium ore contains rutile (TiO2) plus some oxides of iron and silica. When it is heated with

USPshnik [31]

Explanation:

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4 0

3 years ago

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl−+I−→OI−+Cl− T

motikmotik

Answer :

(a) The rate law for the reaction is:

$\text{Rate}=k[OCl^-]^1[I^-]^1$

(b) The value of rate constant is, $60.4M^{-1}s^{-1}$

(c) rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$OCl^-+I^-\rightarrow OI^-+Cl^-$

Rate law expression for the reaction:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

where,

a = order with respect to $OCl^-$

b = order with respect to $I^-$

Expression for rate law for first observation:

$1.36\times 10^{-4}=k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(1)

Expression for rate law for second observation:

$2.72\times 10^{-4}=k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b$ ....(2)

Expression for rate law for third observation:

$2.72\times 10^{-4}=k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(3.0\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}\\\\2=2^a\\a=1$

Dividing 1 from 3, we get:

$\frac{2.72\times 10^{-4}}{1.36\times 10^{-4}}=\frac{k(1.5\times 10^{-3})^a(1.5\times 10^{-3})^b}{k(1.5\times 10^{-3})^a(3.0\times 10^{-3})^b}\\\\2=2^b\\b=1$

Thus, the rate law becomes:

$\text{Rate}=k[OCl^-]^a[I^-]^b$

a = 1 and b = 1

$\text{Rate}=k[OCl^-]^1[I^-]^1$

Now, calculating the value of 'k' (rate constant) by using any expression.

$1.36\times 10^{-4}=k(1.5\times 10^{-3})(1.5\times 10^{-3})$

$k=60.4M^{-1}s^{-1}$

Now we have to calculate the rate for a reaction when concentration of $OCl^-$ and $I^-$ is $1.8\times 10^{-3}M$ and $6.0\times 10^{-4}M$ respectively.

$\text{Rate}=k[OCl^-][I^-]$

$\text{Rate}=(60.4M^{-1}s^{-1})\times (1.8\times 10^{-3}M)(6.0\times 10^{-4}M)$

$\text{Rate}=6.52\times 10^{-5}Ms^{-1}$

Therefore, the rate of the reaction is $6.52\times 10^{-5}Ms^{-1}$

8 0

3 years ago