1 Answers
Call
Talk Doctor Online in Bissoy App
How does a barium atom become a barium ion with a $2^+$ charge?
Basically, an atom becomes an ion when it "steals" electrons from another atom or when another atom "steals" electrons from it. Barium becomes an ion with a $2^+$ charge when...
1 Answers 1 views
The electron configuration $1s^2 2s^2 2p^6 3s^2 3p^6$ represents which noble gas?
The given means that the element has 18 electrons - 2 in the first energy level and in the 2 orbital, 8 in the second energy level of which 2...
1 Answers 1 views
To make barium sulfate from 1 equiv barium nitrate, how many equiv of sodium sulfate are required?
$Na_2SO_4(aq) + Ba(NO_3)_2(aq) rarr BaSO_4(s)darr+ 2NaNO_3(aq)$ Barium sulfate is pretty insoluble stuff, and will precipitate from solution as a white solid. Its solubility in aqueous solution is negligible.
1 Answers 1 views
What would be observed in the addition of barium nitrate to magnesium sulfate? What about for barium nitrate and postassium iodide?
Barium nitrate and magnesium sulfate: $Ba(NO_3)_2(aq) + MgSO_4(aq) rarr BaSO_4(s)darr + Mg(NO_3)_2(aq)$ Barium sulfate is quite insoluble in water. A white solid precipitates. Barium nitrate and potassium iodide: Both metathesis...
1 Answers 1 views
How many grams of barium sulfate solid are produced from reacting 18.48 grams of barium chloride with an excess amount of sodium sulfate?
$BaCl_2(aq) + Na_2SO_4(aq) rarr BaSO_4(s)darr + 2NaCl(aq)$ $"Moles of barium chloride"=(18.48*g)/(208.23*g*mol^-1)=8.87xx10^-2*mol$ Barium chloride is the . At MOST, we can get $8.87xx10^-2*mol$ $BaSO_4(s)$, i.e. $8.87xx10^-2*molxx233.38*g*mol^-1=??g$ $"Barium sulfate"$ is as soluble...
1 Answers 1 views
A 3.060 g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample.The resultant reaction produced a precipitate of barium sulfate, which collected by filtration, washed, dried, weight?
For $"BaSO"_4$, $M_r = "137.33 + 32.06 + 64.00 = 233.39"$ ∴ $0.2745 color(red)(cancel(color(black)("g BaSO"_4))) × "137.33 g Ba"/(233.99 color(red)(cancel(color(black)("g BaSO"_4)))) = "0.1611 g Ba"$ $"% Ba" = ("0.1611" color(red)(cancel(color(black)("g"))))/(3.060...
1 Answers 1 views
A 211 g sample of barium carbonate reacts with a solution of nitric acid to give barium nitrate, carbon dioxide, and water. If the acid is present in excess, what mass and volume of carbon dioxide gas at STP will form?
$BaCO_3(s) + 2HNO_3(aq) rarrBa(NO_3)_2(aq)+ H_2O(l) + CO_2(g)uarr$ Given the , ONE MOLE of carbon dioxide is evolved for each mole of barium carbonate. $"Moles of barium carbonate"=(211*g)/(197.34*g*mol^-1)=1.07*mol.$ Given the...
1 Answers 1 views
A solution of barium hydroxide, $Ba(OH)_2$ contains 4.285 grams of barium hydroxide in 100 mL of solution. What is the molarity of the solution?
We know the molar mass of barium hydroxide is 171.34 (you can google this or just add the mass of each atom in the molecule). Knowing that $molarity = (mols)...
1 Answers 1 views
What mass of barium chloride is present in a $100*mL$ of barium chloride solution of $0.250*mol*L^-1$ concentration?
And thus, $"moles of stuff"="concentration"xx"volume of solution"$. And in turn, $"moles of stuff"="Mass of stuff"/"Molar mass of stuff"$. We have a $100*cm^3$ volume of a $0.250*mol*L^-1$ solution of $BaCl_2$. And...
1 Answers 1 views
A stock solution contains 51.24 g of barium hydroxide in 1.20 L of solution. How do I make 1.00 L of 0.100 mol/L barium hydroxide from the stock solution?
You must dilute 401 mL with enough water to make 1.00 L of solution. Step 1. Calculate the of the stock solution. Moles of Ba(OH)₂ = $"51.24 g Ba(OH)"_2 ×...
1 Answers 1 views