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MissTica
3 years ago
6

Can someone please help me on the second one? I have no idea what to do.

Chemistry
1 answer:
Aleks [24]3 years ago
8 0

The reaction will produce 12.1 g Ag₂S.

<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S

<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)

× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S

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What evidence supports a conservation law? 6 CO2 → C6H12O6 6 O2 6 CO2 6 H2O light → C6H12O6 6 O2 6 H2O light → C6H12O6 6 O2 6 CO
docker41 [41]

The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

  • \rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

Learn more about conservation law, here:

brainly.com/question/2175724

4 0
2 years ago
The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6
s344n2d4d5 [400]

Answer:

the conversion factor is f= 6  mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6  mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

3 0
3 years ago
How does pH affect enzyme-catalyzed reactions? How does affect enzyme-catalyzed reactions? Energy stored in protons is used to d
Zolol [24]

Answer:

The concentration of protons affects an enzyme's folded structure and reactivity.

Explanation:

Enzymes act within narrow pH limits (optimal reaction pH). Since most enzymes have a protein structure, the variation in pH or temperature affects their enzymatic activity.

To catalyze a reaction, an enzyme binds to one or more reagent molecules. These molecules are the substrates of the enzyme.

In some reactions, a substrate breaks into several products. In others, two substrates join together to create a larger molecule or to exchange parts. In fact, for any biological reaction that can occur to you, there is probably an enzyme to accelerate it.

The part of the enzyme where the substrate binds is called the active site.

The amino acid residues of the active site often have acidic or basic properties that are important for catalysis. Changes in pH can affect these residues and make binding with the substrate difficult.

7 0
3 years ago
0.250 moles of NaCl is dissolved in 0.850 L of solution. What is the molar concentration?
Nuetrik [128]

Considering the definition of molarity, the molar concentration is 0.294 \frac{moles}{liter}.

Molarity reflects the concentration of a solution indicating the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{amount of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}.

In this case, you know:

  • amount of moles of solute= 0.250 moles
  • volume= 0.850 L

Replacing in the definition of molarity:

molarity=\frac{0.250 moles}{0.850 L}

Solving:

molarity= 0.294 \frac{moles}{liter}

Finally, the molar concentration is 0.294 \frac{moles}{liter}.

Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults

6 0
3 years ago
Calculate the enthalpy of the reaction
harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)     \Delta H_A=+2035kJ

(2) 2B(s)+3H_2(g)\rightarrow B_2H_6(g)    \Delta H_B=+36kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_C=-285kJ

(4) H_2O(l)\rightarrow H_2O(g)    \Delta H_D=+44kJ

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D

\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)

\Delta H=-2552kJ

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0
3 years ago
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