<span>C2H5
First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2.
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass.
moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles
moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles
The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule.
Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen.
moles C = 0.50899
moles H = 0.638361 * 2 = 1.276722
We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon.
total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185
7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked.
Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen.
0.50899 / 1.276722 = 0.398669
0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5.
Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is
C2H5</span>
Answer:
Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)
Explanation:
Part 1. Volume of reactant
(a) Balanced chemical equation.

(b) Moles of CuCl₂

(c) Moles of Na₃PO₄
The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

(d) Volume of Na₃PO₄

Part 2. Net ionic equation
(a) Molecular equation

(b) Ionic equation
You write molecular formulas for the solids, and you write the soluble ionic substances as ions.
According to the solubility rules, metal phosphates are insoluble.
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)
(c) Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>
The net ionic equation is
3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)
Answer:Butane > ethane > methane, because between bigger molecules there are stronger van der Waals forces and also higher molar mass means they need to be given more energy to have enough kinetic energy to move quickly, freely in gas.
There are multiple butene isomers (Butene) and some (2-Butenes - cis and trans) actually have higher boiling point than n-Butane (there is also Isobutane, of course, with quite much lower boiling point than all of them) and some (1-Butene, Isobutylene) have lower, so this isn't really a fair or simple question. But on simplest level, it can again be said that 1-butene has lower boiling point because it has very similar shape but slightly lower molar mass (2H less) than n-butane.
Explanation:
The neutron, Proton, and Electron
The H+ in a solution that has a Ph of 8.73 is calculate as follows
Ph is always = - log (H+)
H+ = 10^-Ph
H+ is therefore = 10 ^- 8.73
H+ = 1.86 x10^-9 M