Physical change is one that affects the physical properties of a substance true or false

Write the oxidation and reduction half reactions;

luda_lava [24]

Answer:

a)

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

$Br_2+2e^-$ ⇒ $2Br^-$

b)

$Mg$ ⇒ $Mg^{2+}+2e^-$

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

Explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

$Br_2+2e^-$ ⇒ $2Br^-$

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

$Mg$ ⇒ $Mg^{2+}+2e^-$

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

3 0

3 years ago

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If I have 2 mol NaOH in 3 L of solution, what is the molarity of my solution?

qwelly [4]

Answer:

0.67 mole/litre

Explanation:

the molarity equall no. of moles ÷ volume of sol.

3 0

2 years ago

Consider the following scenario

GarryVolchara [31]

Answer:

See explaination

Explanation:

Going by the clues that it is between Silver Flouride (AgF) and Sodium Fluoride (NaF) and since it is an aqueous solution , the 1 liter bottle is likely to be Sodium Chloride( NaCl). Going by the reaction,

AgF + NaCl= AgCl + NaF

Here, the color of AgCl is white, hence the solution cannot be AgCl.

Determination of NaCl

Determination of NaCl can be done by Mohr's Method or Volhard's method. But results in Volhard's method are more accurate . Its uses the method of back titration with Potassium Thiocynate which forms a AgCl precipitate . Prior to titration,excess AgNO3 ( The problem also has a clue that excess reagents are present in the lab ) is added to the NaCl solution so that all the Cl- ions react with Ag+. Fe3+ is then added as an indicator and the solution is titrated with KSCN to form a silver thiocyannite precipitate (AgSCN). Once all the silver has reacted, a slight excess of SCN- reacts with Fe3+ to form Fe(SCN)3 dark red complex. The concentration of Cl- is determined by subtracting the titer findings of Ag+ ions that reacted to form AgSCN from the Ag NO3 moles added to the solution. This is used because pH of the solution is acidic. If the pH of solution is basic, Mohr's method is used.

Reactions

Ag+ (aq)+ Cl-(aq) = AgCl(aq)

Ag+(aq) + SCN-(aq) = AgSCN(aq)

Fe3+(aq) + SCN-(aq) = [FeSCN]2- (aq)

7 0

3 years ago

At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz

Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P° (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles: n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

Hope this helps

7 0

2 years ago

Protons are positively charged and repel other protons. Which other particle is found in the nucleus and separates protons so th

Leto [7]

Neutrons which doesn't have any charge

6 0

3 years ago

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