Answer:
A. In a graduated cylinder, put some quantity of water and measure the initial volume. Then put a coin and measure the volume. To find the volume of the coin, simply subtract the initial volume (water only) from the ending volume (water + coin). To measure the mass, take a dry coin and place it on an electronic scale. Density = mass / volume, so divide the mass by the volume to calculate the density of the coin.
B. When measuring the volume, make sure to look at the graduated cylinder at eye level and read from the bottom of the meniscus.
Moles of phosphoric acid would be needed : 0.833
<h3>Further explanation</h3>
Given
15 grams of water
Required
moles of phosphoric acid
Solution
Reaction(decomposition) :
H3PO4 -> H2O + HPO3
mol water (H2O :
= mass : MW
= 15 g : 18 g/mol
= 0.833
From the equation, mol ratio H3PO4 = mol H2O = 1 : 1, so mol H3PO4 = 0.833
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.