Fertile offspring mate is a form of reproduction
In order to <span>decrease the pressure of a gas inside a closed cubical container, you need to decrease the temperature of the container. The volume of the system is rigid so it means volume is constant. By the ideal gas law, temperature and pressure are directly related. Increasing the temperature, increases the pressure and the opposite to happens.</span>
Answer:
21.10g of H2O
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2C7H14 + 21O2 —> 14CO2 + 14H2O
From the balanced equation above, 2L of C7H14 produced 14L of H2O.
Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.
Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:
1 mole of a gas occupy 22.4L at stp
Therefore, Xmol of H2O will occupy
26.25L i.e
Xmol of H2O = 26.25/22.4
Xmol of H2O = 1.172 mole
Therefore, 1.172 mole of H2O is produced from the reaction.
Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:
Number of mole H2O = 1.172 mole
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O =..?
Mass = mole x molar mass
Mass of H2O = 1.172 x 18
Mass of H2O = 21.10g
Therefore, 21.10g of H2O is produced from the reaction.
A. Oxygen. Oxygen has an atomic mass of 16. The atomic mass of an atom is the combined weight o the protons and neutrons. Since Oxygen's atomic mass is 16, it has 8 protons and 8 neutrons.
Answer: The mass percentage of
is 5.86%
Explanation:
To calculate the mass percentage of
in the sample it is necessary to know the mass of the solute (
in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).
To calculate the mass of the solute, we must take the mass of the
precipitate. We can establish a relation between the mass of
and
using the stoichiometry of the compounds:

Since for every mole of Tl in
there are two moles of Tl in
, we have:

Using the molar mass of
we have:

Finally, we can use the mass percentage formula:
