Why does fluorine (F) have the strongest attraction between its nucleus and its valence electrons?

The reaction Ba(NO3)2(aq) + Na2SO4(aq) >> 2NaNO3(aq) + BaSO4(s) goes to completion because a

Helen [10]

Answer:

2. precipitate is formed

Explanation:

In this reaction barium nitrate and sodium sulfate solutions react to form a barium sulfate precipitate.Barium and sulfate ions will react to give barium sulfate precipitate where as the sodium and nitrate ions are spectator ions.

The net ionic equation after removing unchanged ions from each side of the full ionic equation will be;

Ba²⁺+ SO₄²⁻ → BaSO₄ (s)

3 0

3 years ago

Read 2 more answers

Gaseous ethane (CH,CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO) and gaseous water (H2O). Supp

notka56 [123]

Answer:

0.00 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ 30.07 32.00

2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g: 1.50 11.

2. Calculate the moles of each reactant

$\text{moles of C$_{2}$H}_{6} = \text{1.50 g C$_{2}$H}_{6} \times \dfrac{\text{1 mol C$_{2}$H}_{6}}{\text{30.07 g C$_{2}$H}_{6}} = \text{0.04988 mol C$_{2}$H}_{6}\\\\\text{moles of O}_{2} = \text{11. g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.34 mol O}_{2}$

3. Calculate the moles of CO₂ we can obtain from each reactant

From ethane:

The molar ratio is 4 mol CO₂:2 mol C₂H₆

$\text{Moles of CO}_{2} = \text{0.04988 mol C$_{2}$H}_{6} \times \dfrac{\text{4 mol CO}_{2}}{\text{2 mol C$_{2}$H}_{6}} = \text{0.09976 mol CO}_{2}$

From oxygen:

The molar ratio is 4 mol CO₂:7 mol O₂

$\text{Moles of CO}_{2} = \text{0.34 mol O}_{2}\times \dfrac{\text{4 mol CO}_{2}}{\text{7 mol O}_{2}} = \text{0.20 mol CO}_{2}$

4. Identify the limiting and excess reactants

The limiting reactant is ethane, because it gives the smaller amount of CO₂.

The excess reactant is oxygen.

5. Mass of ethane left over.

Ethane is the limiting reactant. It will be completely used up.

The mass of ethane left over will be 0.00 g.

8 0

3 years ago

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should as

FrozenT [24]

Answer:

0.100 M AlCl₃

Explanation:

The variation of boiling point by the addition of a nonvolatile solute is called ebullioscopy, and the temperature variation is calculated by:

ΔT = W.i

Where W = nsolute/msolvent, and i is the Van't Hoff factor. Because all the substances have the same molarity, n is equal for all of them.

i = final particles/initial particles

C₆H₁₂O₆ don't dissociate, so final particles = initial particles => i = 1;

AlCl₃ dissociates at Al⁺³ and 3Cl⁻, so has 4 final particles and 1 initial particle, i = 4/1 = 4;

NaCl dissociates at Na⁺ and Cl⁻ so has 2 final particles and 1 initial particle, i = 2/1 = 2;

MgCl₂ dissociates at Mg⁺² and 2Cl⁻, so has 3 final particles and 1 initial particle, i = 3/1 = 3.

So, the solution with AlCl₃ will have the highest ΔT, and because of that the highest boiling point.

8 0

4 years ago

Consider the nuclear equation below. 15/8 O ==> 15/7 N + 0/+1 e Which correctly identifies the type of reaction?

sweet [91]

B: Beta plus Decay :)

6 0

3 years ago

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.60 m K3PO4(aq).

Serga [27]

To find the freezing point, you use the following formula

fp solution= fp solvent - Δfp solution

fp solvent= 0 C (since the solvent was not stated, you can assume is water which freezing point is 0 degree celsius)

but first, we need to find the depression in freezing point (Δfp solution)

Δfp solution= Kf x i x m, where Kf is constant for the solvent, i is the number of particles produced by the solute and m is the molality of the solution.

Kf= 1.86 for water
i= 4 , becausethe solute breaks into 4 ions, K3PO4---> 3K+ + PO4-3
m= 2.60

Δfp solution= (1.86) (4) (2.60)= 19.3 C

fp solution= 0 - 19.3C=<span> -19.3C
</span>

for boiling point, we use a similar formula and way to solve it.

Δbp solution= bp solvent + Δbp solution

bp solvent= 100 C for water

to find the elevation boiling point (Δbp soln) is equal to:

Δbp solution= Kb x i x m

Kb= 0.512
i= 4
m= 2.60

Δbp solution= (0.512) (4) (2.60)= 5.32 C

bp solution= 100C + 5.32C= <span>105.32 C</span>

6 0

4 years ago