Which sub atomic particles are similar in size

463.5 grams of HCl are required to react with 6.35 moles of CaCO₃.

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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The molecular formula of the compound, given that it contains 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen is C₆H₁₂N₄

To obtain the molecular formula, we must first determine the empirical formula. Details on how to obtain the empirical formula is given beloww:

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

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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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.

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.

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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The mass of calcium in the original urine sample would be 0.0140 g.

First, 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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