Answer:
2.5g
Explanation:
When the reaction goes into completion, they will produce 2.5g. This is complement the law of conservation of mass.
According to the law of conservation of mass "in a chemical reaction, matter is neither created nor destroyed but transformed from one form to another".
- The mass of reactants and products in a chemical reaction must be the same.
- There is no change in mass in moving from reactant to product
- So, if we start with 2.5g of reactants, we must end with 2.5g of products.
Suppose we have 100 gr of the substance. Then by weight, it would contain 44.77 gr of C, 7.46 gr of H and 47.76 gr of S. We need to look up the atomic weights of these atoms; M_H=1, M_C=12, M_S=32. The following formula holds (where n are the moles of the substance, M its molecular mass and m its mass): n=m/M. Substituting the known quantities for each element, we get that the substance has 3.73 moles of C, 7.46 moles of H and 1.49 moles of S. In the empirical formula for the molecule, all atoms appear an integer amout of times. Hence, for every mole of Sulfur, we have 2.5 moles of C and 5 moles of H (by taking the moles ratios). Thus, for every 2 moles of sulfur, we have 5 moles of C and 10 moles of H. Now that all the coefficients are integer, we have arrived at an empirical formula for the skunk spray agent:
Answer:
<h2>A star's life cycle is determined by its mass. The larger its mass, the shorter its life cycle. A star's mass is determined by the amount of matter that is available in its nebula, the giant cloud of gas and dust from which it was born. ... As the gas spins faster, it heats up and becomes as a protostar.</h2>
Explanation:
<h2>Read this and then choose your options ✍️✍️</h2>
Answer
321.8 g CaF2
321.5 g Al2(CO3)3
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.