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Gnom [1K]
3 years ago
11

What is the stoichiometric ratio between BaCl2 and NaCl

Chemistry
1 answer:
bixtya [17]3 years ago
7 0
<span>BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
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Answer:

36000 centimeters

Explanation:

If the average speed of a species of tortoise is 0.36 km/h, then it would be 36000 centimeters.

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3 years ago
What are the three main particles in the modern model of an atom
EastWind [94]
Protons and neutrons in the nucleus (the bit in the middle) and small tiny electrons that orbit the nucleus
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2 years ago
What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks
Firlakuza [10]

Answer:

\large \boxed{1.64\times 10^{-5}\text{ mol/L }}

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}

 

3 0
3 years ago
Abigail obtained 36.6 grams of calcium carbonate after performing a reaction. From her calculations, she knew she should have ob
Zielflug [23.3K]

Answer:

it would be 1

Explanation:

3 0
3 years ago
Select all of the following statements that are false about ΔGo and ΔG:a) If the reaction has a large negative ΔGo value, the re
OLEGan [10]

Answer:

a) If the reaction has a large negative ΔGo value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔGo and ΔG have the same magnitude, they just have opposite signs.

Explanation:

The fraction of the total heat energy if a system that does useful work is known as Gibb's free energy (G) and the change from the initial to final state is designated by \Delta G. It is observed that the values of \Delta G changes with experimental conditions such as temperature , pressure , concentration etc.

\Delta G^0 is the standard free energy change which is a balance of two natural tendencies of any system.

  1. Minimization of potential energy or enthalpic factor \Delta H^0

Maximization of disorderliness or entropic factor T\Delta S^0

Mathematically; \Delta G = \Delta H^0 - T\Delta S^0

Thus; from above mentioned, the statements that are true about ΔG⁰ and ΔG are:

ΔG⁰ and ΔG can have different values, they don't even have to have the same sign

For a reaction that reaches equilibrium, the minimum value of free energy must be at the equilibrium point

If ΔG⁰ , measured at an extent of reaction = 0.5, is positive, the sign for ΔG when the extent of reaction = 0.80 is also positive.

while the false statements include:

a) If the reaction has a large negative ΔG⁰ value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔG⁰ and ΔG have the same magnitude, they just have opposite signs.

7 0
3 years ago
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