All categories

2 years ago

A student made a graph to show the chemical equilibrium position of a reaction.

Chemistry

1 answer:

pogonyaev2 years ago

8 0

Answer:

Concentration, because the amounts of reactants and products remain constant after equilibrium is reached.

Explanation:

The rate of reaction refers to the amount of reactants converted or products formed per unit time.

As the reaction progresses, reactions are converted into products. This continues until equilibrium is attained in a closed system.

When equilibrium is attained, the rate of forward reaction is equal to the rate of reverse reaction, hence the concentration of reactants and products in the system remain fairly constant over time.

When deducing the rate of reaction, concentration of the specie of interest is plotted on the y-axis against time on the x-axis.

You might be interested in

Calculate the molarity of an HCl solution if 23.88 mL of it reacts with 6.5287 grams of sodium carbonate (106 g/mol) according t

Ivanshal [37]

Answer:

5.158 mol/L

Explanation:

To find the molarity, you need to use the formula:

Molarity (M) = moles / volume (L)

You have been grams sodium carbonate. You need to (1) convert grams Na₂CO₃ to moles (via molar mass), then (2) convert moles Na₂CO₃ to moles HCl (via mole-to-mole ratio from equation), then (3) convert mL to L (by dividing by 1,000), and then (4) use the molarity equation.

Steps 1 - 2:

2 HCl + 1 Na₂CO₃ ----> 2 NaCl + H₂O + CO₂

6.5287 g Na₂CO₃ 1 mole 2 moles HCl
-------------------------- x ------------- x ------------------------- = 0.12318 mole HCl
106 g 1 mole Na₂CO₃

Step 3:

23.88 mL / 1,000 = 0.02388 L

Step 4:

Molarity = moles / volume

Molarity = 0.12318 mole / 0.02388 L

Molarity = 5.158 mole/L

**mole/L is equal to M**

7 0

1 year ago

How do you solve for the Atomic Mass in chemistry?

loris [4]

Https://www.google.com/search?q=how+to+solve+fir+atomic+mass+in+chemisty&ie=UTF-8&oe=UTF-8&hl=en-us&client=safari#kpvalbx=1
Here is the link to a great video that explains your question nicely, hope this helps.

6 0

3 years ago

Read 2 more answers

How many atoms are in 6.4 moles of vanadium

den301095 [7]

6.4 * 6.02 * 10^23 = 3.8528*10^24 atoms
Don't let the fact that it's vanadium throw you off, avagadros constant stays the same for all elements

3 0

3 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$

To calculate the $E^o_{(Cu^{2+}/Cu)}$ of the reaction, we use the equation:

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

$E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}$

Putting values in above equation, we get:

$1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)$

$E^o_{(Cu^{2+}/Cu)}=0.34V$

Hence, the value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

8 0

2 years ago

A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. the boiling point of this solution was dete

Katarina [22]

We will use boiling point formula:

ΔT = i Kb m

when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35

and Kb is the boiling point constant =5.03

and m = molality

i = vant's Hoff factor

so by substitution, we can get the molality:

1.35 = 1 * 5.03 * m

∴ m = 0.27

when molality = moles / mass Kg

0.27 = moles / 0.015Kg

∴ moles = 0.00405 moles

∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol

5 0

3 years ago