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2 years ago

5

Think of the different types of reactions you have learned about. How can you tell if they absorb energy or release energy, and

at the molecular level, how are they different?

1 answer:

Leokris [45]2 years ago

6 0

Answer:

well the reactions are probably wow and amazed

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How many sodium atoms will create an ionic bond with sulfur?

Galina-37 [17]

Hi,

Two sodium atoms are needed to create an ionic bond with sulfur.

5 0

3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago

What type of physical weathering is caused by the expansion of water?

skelet666 [1.2K]

Answer:

wedging

Explanation:

ice or frost wedging

6 0

3 years ago

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

$Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)$

Reduction at cathode :

$Cu^+(aq)+2e^-\rightarrow Cu(s)$

Reduction potential of $Cu^{2+}$ to Cu= $E^o_{1}=0.34 V$

Oxidation at anode:

$Fe(s)\rightarrow Fe^{2+}(aq)+2e^-$

Reduction potential of $Fe^{2+}$ to Fe= $E^o_{2}=-0.44 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.34V -(-0.44 V)=0.78 V$

The standard cell potential of the reaction is 0.78 Volts.

3 0

3 years ago

Which technique would be best for separating sand and water?

ipn [44]

Answer:

A. filtration

Hope it helps

3 0

3 years ago