76.33 grams of NaCl were collected after experiment 1.306 mol were
produced.
What is mole formula?Every material has a molecular weight of 6.023 x 10²³. It may be used to quantify the chemical reaction's byproducts. The symbol mol is used to identify the unit. The molecular formula is written out as follows.
Mass of material / mass of one mole equals the number of moles.
We need to know the molar mass of NaCl in order to compute the number of moles of NaCl created.
The atomic weights of sodium (Na) and chlorine together make up the molar mass of sodium chloride (Cl). Na has an atomic mass of 22.99 g/mol, while Cl has an atomic mass of 35.45 g/mol. As a result, NaCl's molar mass is:
Molar mass of NaCl
= (1 x atomic mass of Na) + (1 x atomic mass of Cl)
= (1 × 35.45 g/mol plus 1 x 22.99 g/mol)
= 58.44 g/mol
The mass of gathered NaCl may now be converted into moles using the molar mass:
Mass of NaCl divided by its molar mass yields moles of NaCl.
moles of NaCl = 76.33 g / 58.44 g/mol
moles of NaCl = 1.306 mol
As a result, the experiment generated 1.306 moles of NaCl.
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consider the multistep reaction below. what is the balanced chemical equation of the overall reaction?
The overall reaction of the multistep reaction is: 2A + B → C + D
This reaction can be broken down into two individual steps. In the first step, A and B react to form an intermediate product, X. The balanced chemical equation for this step is: A + B → X. In the second step, the intermediate product X is reacted with A to form C and D. The balanced chemical equation for this step is:X + A → C + D
Combining these two equations yields the overall balanced chemical equation:
2A + B → C + D
In summary, the overall balanced chemical equation for the multistep reaction is 2A + B → C + D. This equation shows that two molecules of A and one molecule of B will combine to form one molecule of C and one molecule of D.
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the amount of kinetic energy required to strain the chemical bonds in substrates so they can achieve the transition state is the definition of ?
The amount of kinetic energy required to strain the chemical bonds in substrates so they can achieve the transition state is the definition of activation energy.
What is Activation Energy?
Activation energy is the amount of energy required for a chemical reaction to occur. The energy that must be provided to molecules in order for them to react with one another is known as activation energy.
This can be accomplished in a variety of ways, such as by increasing the temperature or pressure, adding a catalyst, or irradiating the reactants with light.
Activation energy is defined as the energy required for the reaction to begin. It's the energy that molecules require to overcome the initial barrier so that a reaction may proceed.
When a chemical reaction occurs, the reactants must collide with one another with sufficient force and in the appropriate orientation to form products.
It's critical to note that activation energy is a form of potential energy that isn't included in the overall energy change of a reaction.
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How many grams of aluminum sulfate would be formed if 3.52 grams of aluminum completely reacted with H2SO4?
2Al + 3H2SO4 ---------------------> Al2(SO4)3 + 3H2
Taking into account the reaction stoichiometry, 22.29 grams of Al₂(SO₄)₃ are formed if 3.52 grams of aluminum completely reacted with H₂SO₄.
Reaction stoichiometryThe balanced reaction is:
2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂
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:
Al: 2 molesH₂SO₄: 3 molesAl₂(SO₄)₃: 1 molH₂: 3 molesThe molar mass of the compounds is:
Al: 27 g/moleH₂SO₄: 98 g/moleAl₂(SO₄)₃: 342 g/moleH₂: 2 g/moleBy reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Al: 2 moles ×27 g/mole= 54 gramsH₂SO₄: 3 moles ×98 g/mole= 294 gramsAl₂(SO₄)₃: 1 mol ×342 g/mole= 342 gramsH₂: 3 moles ×2 g/mole= 6 gramsMass of Al₂(SO₄)₃ formedThe following rule of three can be applied: if by reaction stoichiometry 54 grams of Al form 342 grams of Al₂(SO₄)₃, 3.52 grams of Al form how much mass of Al₂(SO₄)₃?
mass of Al₂(SO₄)₃= (3.52 grams of Al× 342 grams of Al₂(SO₄)₃)÷ 54 grams of Al
mass of Al₂(SO₄)₃= 22.29 grams
Finally, 22.29 grams of Al₂(SO₄)₃ are formed.
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it is found that, when equilibrium is reached at a certain temperature, hi is 40. percent dissociated. calculate the equilibrium constant kc for the reaction at this temperature.
The equilibrium constant (Kc) is the ratio of the concentration of the products to the reactants at equilibrium. The value of Kc changes with the temperature but is constant at a given temperature.
The expression for the equilibrium constant Kc can be defined as follows:-
Kc = [C]^c[D]^d/[A]^a[B]^b
where [ ] denotes the molar concentration of the respective species. a, b, c, and d are the coefficients of the balanced chemical equation for the species A, B, C, and D.
If a chemical reaction is at equilibrium at a given temperature, the concentration of reactants and products remains constant over time. In other words, the rate of the forward reaction and the rate of the reverse reaction is equal.
The reaction for which we need to find the equilibrium constant is:-
HI(g) ↔ H(g) + I(g)
Now, assume that initially there were 'x' moles of HI in the reaction mixture. After the dissociation of HI, the concentration of H and I will be equal to 'x - y' moles. The concentration of HI will be equal to 'x - y' moles.
Here, y is the number of moles of HI that dissociated. According to the given statement, HI is 40% dissociated. Therefore, the number of moles of HI that dissociated will be 0.4x. Similarly, the number of moles of H and I that will be formed will also be 0.4x.
The equation for the dissociation of HI can be written as:-
HI(g) ↔ H(g) + I(g)
The initial number of moles = x Moles dissociated = 0.4x
At equilibrium, the number of moles of HI = x - 0.4x = 0.6x
Number of moles of H = 0.4x
Number of moles of I = 0.4x
Finally, substitute these values in the expression for the equilibrium constant:-
Kc = [H][I]/[HI]
Kc = (0.4x)(0.4x)/(0.6x)²
Kc = 0.16/0.36Kc = 0.4444 (approximately)
Therefore, the equilibrium constant Kc for the given reaction is 0.4444 (approximately).
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