Yes, If all the coefficients in a balanced chemical equation are multiplied by 2, the equation will still remain balanced because the ratio of the reactants and products remain the same.
What are the coefficients?
For example, the balanced equation: [tex]2H_{2}[/tex] + [tex]O_{2}[/tex] -> [tex]2H_{2}O[/tex]
The coefficients of a chemical equation are the numbers written in front of the chemical formulas of reactants and products, indicating the relative amounts of each substance involved in the reaction. The coefficients are used to balance the chemical equation, ensuring that the law of conservation of mass is obeyed. The coefficients represent the smallest whole-number ratios of the substances in the reaction, and they provide important information about the stoichiometry of the reaction.
When all the coefficients are multiplied by 2, the equation becomes: [tex]4H_{2}[/tex] + [tex]2O_{2}[/tex] -> [tex]4H_{2}O[/tex]
The equation is still balanced because the ratio of hydrogen to oxygen to water molecules remains the same (4:2:4).
Multiplying the coefficients by a constant will not affect the equilibrium constant (Kc) as long as the reaction conditions remain constant. This is because the equilibrium constant is a ratio of the concentrations of products and reactants at equilibrium, and the ratio remains the same even if the coefficients of the balanced equation are multiplied by a constant. However, if the temperature, pressure or concentration of any reactants or products are changed, then the value of the equilibrium constant will change.
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The two possible units of molarity are
Answer: The units for molarity are moles/liter.
Similarly, the equation to find molarity is moles divided by liters.
Explanation:
mol / L is a unit of molar concentration. These are the number of moles of dissolved material per liter of solution. 1 mol / L is also called 1M or 1molar. Mol / m3 is also a unit of molar concentration.
Molarity is expressed in units of moles per liter (mol / L). This is a very common unit, so it has its own symbol, which is the uppercase M. A solution with a concentration of 5 mmol / l is called a 5 M solution or has a concentration value of 5 mol.
The molar concentration of the solution is equal to the number of moles of the solute divided by the mass of the solvent (kilogram), and the molar concentration of the solution is equal to the number of moles of the solute divided by the volume of the solution (liter). increase.
What is true of spontaneous reactions?
O They are indicated by a negative change in Gibbs free energy.
O They have a positive value of AS.
O They are instantaneous.
O They always release heat.
Help 20pts
The calcium and magnesium in a urine sample were precipitated as oxalates. A mixed precipitate of calcium oxalate (CaC2O4) and magnesium oxalate (MgC2O4) resulted and was analysed by gravimetry. The formed precipitate mixture was heated to form calcium carbonate (CaCO3) and magnesium oxide (MgO) with a total mass of 0.0433 g. The solid precipitate mixture was ignited to form CaO and MgO, the resulting solid after ignition weighed 0.0285 g. What was the mass of calcium in the original sample? All answers should be reported with the correct significant figures
The mass of calcium in the original urine sample would be 0.0140 g.
Stoichiometric problemFirst, we need to find the masses of calcium and magnesium oxalates in the original sample. Let x be the mass of calcium oxalate and y be the mass of magnesium oxalate. Then we have:
x + y = mass of the mixed oxalate precipitate
Next, we need to use the information given to find the mass of calcium in the original sample. The mass of calcium oxide formed after ignition is equal to the mass of calcium oxalate in the original sample. We can calculate the mass of calcium oxide using the mass of calcium carbonate formed and the molar mass ratio of calcium carbonate to calcium oxide.
The balanced chemical equations for the reactions are:
CaC2O4 -> CaCO3 + CO2
CaCO3 -> CaO + CO2
The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of CaO is 56.08 g/mol.
From the given information, we have:
0.0433 g = (x + y)(100.09 g/mol + 80.15 g/mol) / (128.10 g/mol + 80.15 g/mol)
0.0285 g = x(56.08 g/mol) + y(40.31 g/mol)
Solving these equations simultaneously, we get:
x = 0.0140 g
y = 0.0053 g
Therefore, the mass of calcium in the original sample (which is equal to the mass of calcium oxide formed after ignition) is:
0.0140 g
So the mass of calcium in the original sample is 0.0140 g.
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CaCO3 + 2HCI =CaCl2 + H₂O + CO2
5. Calcium carbonate (CaCO3) combines with HCl to produce calcium chloride (CaCl₂),
water, and carbon dioxide gas (CO₂). How many grams of HCI are required to react with
6.35 mol CaCO3?
463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.
What is meant by molar mass?Mass of one mole of substance is referred to as the molar mass. The molar mass of a substance can be calculated by adding up the atomic masses of all the atoms in a molecule.
Balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
6.35 mol CaCO₃ * 2 mol HCl / 1 mol CaCO₃ = 12.7 mol HCl
Now, we use the molar mass of HCl (36.46 g/mol) to convert from moles to grams: 12.7 mol HCl * 36.46 g/mol = 463.5 g HCl
Therefore, 463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.
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pls help!!!
a compound is found to be 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen. it has a molecular molar mass of 140.22 g/mol. what is the molecular formula.
show work pls!!
The molecular formula of the compound, given that it contains 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen is C₆H₁₂N₄
How do i determine the molecular formula?To obtain the molecular formula, we must first determine the empirical formula. Details on how to obtain the empirical formula is given beloww:
Carbon (C) = 51.39%Hydrogen (H) = 8.64%Nitrogen (N) = 39.97%Empirical formula =?Divide by their molar mass
C = 51.39 / 12 = 4.283
H = 8.64 / 1 = 8.64
N = 39.97 / 14 = 2.855
Divide by the smallest
C = 4.283 / 2.855 = 1.5
H = 8.64 / 2.855 = 3
N = 2.855 / 2.855 = 1
Multiply through by 2 to express in whole number
C = 1.5 × 2 = 3
H = 3 × 2 = 6
N = 1 × 2 = 2
Thus, we can conclude that the empirical formula is C₃H₆N₂
Finally, we shall determine the molecular formula. Details below
Empirical formula = C₃H₆N₂Molar mass of compound = 140.22 g/molMolecular formula =?Molecular formula = empirical × n = mass number
[C₃H₆N₂]n = 140.22
[(12×3) + (1×6) + (14×2)]n = 140.22
70n = 140.22
Divide both sides by 70
n = 140.22 / 70
n = 2
Molecular formula = [C₃H₆N₂]n
Molecular formula = [C₃H₆N₂]₂
Molecular formula = C₆H₁₂N₄
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A flask filled to the 25.0 ml mark contain 29.97 g of a concentrated salt water solution. What is the density of the solution?
A concentrated saltwater solution weighing 29.97 g and fitting into a flask to the mark of 25.0 ml has a density of about 1199.2 g/L.
How is the density of the solution determined?By dividing the solution's mass by its volume, we may get its density: density = mass/volume
We need to know the density of water at the solution's temperature as well as the capacity of the flask up to the 25.0 ml level in order to calculate the volume of the solution.
Since 1 mL = 0.001 L, volume is equal to 25.0 mL, or 0.0250 L.
Now, we may determine the solution's density as follows:
1199.2 g/L or 29.97 g/0.0250 L is what is referred to as density.
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