Answer: Molarity is defined as moles of solute per liter of solution. So, find the moles of solute and divide by the liters of solution.
molar mass AlCl3 = 133g/mole
moles AlCl3 = 127 g x 1 mole/133 g = 0.955 moles
liters of solution = 400 ml x 1 liter/1000 ml = 0.400 liters
Molarity = 0.955 moles/0.400 liters = 2.39 M
Explain: I looked it up on wyzant.com
Answer:
Kyanite (Al2SiO5) - silicate
Ilmenite (FeTiO3) - Oxides
Rhodochrosite (MnCO3) - carbonate
Celestite (SrSO4) - sulphate
Chalcocite (Cu2S) - sulphide
Explanation:
Minerals are classified according to their chemical composition. For example those that hve the CO32- ion are called carbonates and those with the SO42- ion are called sulphates while the ones with S2- ion are called sulphides
The answer is more protons than electrons.
Answer:
a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
b. 0.957 g
Explanation:
Step 1: Write the balanced equation
2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
Step 2: Convert 130.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15
K = 130.0°C + 273.15
K = 403.2 K
Step 3: Calculate the moles of O₂
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K
n = 2.21 × 10⁻³ mol
Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂
The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.
Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO
The molar mass of HgO is 216.59 g/mol.
4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g
Answer : The volume of gas will be 29.6 L
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 12 atm
= final pressure of gas = 14 atm
= initial volume of gas = 23 L
= final volume of gas = ?
= initial temperature of gas = 200K
= final temperature of gas = 300K
Now put all the given values in the above equation, we get the final pressure of gas.
Therefore, the new volume of gas will be 29.6 L
