10. The lithium-ion in a lithium-ion battery moves from one side to the other by

A chemist adds of a aluminum chloride solution to a reaction flask. Calculate the mass in grams of aluminum chloride the chemist

andreyandreev [35.5K]

Answer:

10.8 g

Explanation:

A chemist adds 480.0 mL of a 0.169 mol/L aluminum chloride solution to a reaction flask. Calculate the mass in grams of aluminum chloride the chemist has added to the flask. Round your answer to 3 significant digits.

Step 1: Given data

Volume of the solution: 480.0 mL

Concentration of the aluminum chloride solution: 0.169 mol/L

Step 2: Calculate the moles of aluminum chloride in the solution

We will multiply the volume of the solution by the molarity.

0.4800 L × 0.169 mol/L = 0.0811 mol

Step 3: Calculate the mass corresponding to 0.0811 moles of AlCl₃

The molar mass of AlCl₃ is 133.34 g/mol.

0.0811 mol × 133.34 g/mol = 10.8 g

3 0

3 years ago

Are Noble gases considered inert gases

Degger [83]

yes because they are inert or inactive at normal temperature, they are inactive because their outer most orbit contain either 2 or 8 electrons.that make them stable

5 0

3 years ago

Read 2 more answers

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com

Elanso [62]

Answer:

(a) $Ksp=4.50x10^{-7}$

(b) $Ksp=1.55x10^{-6}$

(c) $Ksp=2.27x10^{-12}$

(d) $Ksp=1.05x10^{-22}$

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) $BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)$

$Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}$

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

$Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}$

(B) $Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)$

$Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}$

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

$Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}$

(C) $NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)$

$Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}$

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

$Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}$

(D) $La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)$

$Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}$

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

$Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}$

Best regards.

7 0

4 years ago

What element contains the electron configuration 1s22s22p63s23p64s23d104p65s24d8 ?

RideAnS [48]

For better representation, the electronic configuration is written as:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d⁸

The letters s, p, d and f are subshells while the number coefficients represent the energy levels. Each subshell holds a specific maximum number of electrons, which is represented by the exponents. Thus, you add the exponents to determine the atomic number:

Atomic number = 2+2+6+2+6+2+10+6+2+8 = 46

From the periodic table, element 46 is Palladium or Pd.

4 0

3 years ago

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1. Calculate the number of moles of oxygen atoms present in 1.50 mol of barium sulfate.

pochemuha

Answer:

$n_O=6.00molO$

Explanation:

Hello.

In this case, since the molecular formula of barium sulfate is:

$BaSO_4$

Which has four oxygen atoms, we can also say that one mole of barium sulfate has four moles of oxygen; in such a way, the moles of oxygen atoms in 1.50 moles of barium sulfate are:

$n_O=1.50molBaSO_4*\frac{4molO}{1molBaSO_4} \\\\n_O=6.00molO$

Best regards.

3 0

3 years ago