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3 years ago

In the reaction below, how would adding more of reactant A offect the equilibrium of the system? (2 points)

Chemistry

1 answer:

WARRIOR [948]3 years ago

8 0

Answer:

Shift it to the right toward the products,

Explanation:

Adding more reactant will increase the concentration of the reactants, to offset this increased concentration, more products will be formed.

An increase in concentration of a specie favors the direction that uses up that specie and lowers its concentration.
If the concentration of the species on the LHS is increased , the equilibrium will shift to the right to maintain the constancy of the equilibrium constant.

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A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

<u>Answer:</u> The specific heat of metal is 0.821 J/g°C

<u>Explanation:</u>

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C

8 0

3 years ago

Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3

Margarita [4]

<u>Answer:</u> The rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:

$2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O$

The intermediate reaction of the mechanism follows:

<u>Step 1:</u> $H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}$

<u>Step 2:</u> $H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}$

<u>Step 3:</u> $HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[H_3O_2^+][Br^-]$ ......(1)

As, $[H_3O_2^+]$ is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for $[H_3O_2^+]$ from step 1, we get:

$K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}$

$[H_3O_2^+]=K[H^+][H_2O_2]$

Putting the value of $[H_3O_2^+]$ in equation 1, we get:

$\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]$

Hence, the rate law for the reaction is $\text{Rate}=k'[H+][H_2O_2][Br^-]$

4 0

3 years ago

Sound is an example of<br><br> kinetic energy<br> potential energy

motikmotik

Answer-kinetic energy. The human experience of sound is caused by vibrations. The object creating the sound creates waves of movement through a medium, like air, until it reaches our eardrums, which then vibrate and our brain interprets that as sound. Here are some examples of sound energy:

7 0

2 years ago

Read 2 more answers

N the molecule below, how many atoms could make hydrogen bonds with water? the compound has a ch2, double bond, chc, double bond

Anna35 [415]

<h3>Answer:</h3>

10 Atoms

<h3>Explanation:</h3>

The structure of said compound is sketched according to guide lines provided in statement and is attached below.

Hydrogen Bond Interactions:

Hydrogen Bond Interactions are those interactions which are formed between a partial positive hydrogen atom bonded directly to most electronegative atom (i.e. F, O and N) of one molecule and the partial negative most electronegative atom of another molecule.

In given structure we are having seven most electronegative oxygen atoms (labelled red) and three partial positive hydrogen atoms (labelled blue) directly attached to most electronegative atom (i.e. oxygen atoms).

Therefore, the oxygen atoms will make hydrogen bonds with water's hydrogen atoms and the partial positive hydrogen atoms will make hydrogen bonds with water's oxygen atoms respectively.

3 0

3 years ago

In the reaction of 675.9 grams of barium chloride and excess silver(I) nitrate, how many grams of silver(I) chloride should get

zvonat [6]

Taking into account the reaction stoichiometry, the mass of silver(I) chloride formed is 930.37 grams.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + BaCl₂ → 2 AgCl + Ba(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

AgNO₃: 2 moles
BaCl₂: 1 mole
AgCl: 2 moles
Ba(NO₃)₂: 1 mole

The molar mass of the compounds is:

AgNO₃: 169.87 g/mole
BaCl₂: 208.24 g/mole
AgCl: 143.32 g/mole
Ba(NO₃)₂: 261.34 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

AgNO₃: 2 moles ×169.87 g/mole= 339.74 grams

BaCl₂: 1 mole ×208.24 g/mole= 208.24 grams

AgCl: 2 moles ×143.32 g/mole= 286.64 grams

Ba(NO₃)₂: 1 mole×261.34 g/mole= 261.34 grams

<h3>Mass of silver(I) chloride formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 208.24 grams of barium chloride form 286.64 grams of silver(I) chloride, 675.9 grams of barium chloride form how much mass of silver(I) chloride?

$mass of silver(I) chloride=\frac{675.9 grams of barium chloride x286.64 grams of silver(I) chloride }{208.24 grams of barium chloride}$

mass of silver(I) chloride= 930.37 grams

Finally, the mass of silver(I) chloride formed is 930.37 grams.

Learn more about the reaction stoichiometry:

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8 0

1 year ago