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Nitella [24]
3 years ago
7

Calculate the number of moles for 12.044*10^23 number of helium atoms

Chemistry
1 answer:
givi [52]3 years ago
8 0

Answer:

Explanation:

number of moles=number of particle (atoms)/avogadro number (NA)

we know that avogadro number is equal to 6.23*10^23

given number of atoms=12.044*10^23

therefore

moles=12.044*10^23/6.23*10^23=1.9=2

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Consider a solution that is 1.3×10−2 M in Ba2+ and 2.0×10−2 M in Ca2+. Ksp(BaSO4)=1.07×10−10 Ksp(CaSO4)=7.10×10−5 If sodium sulf
skad [1K]

Answer:

Explanation:

Ksp(BaSO4)=1.07×10−10

BaSO₄ → Ba²⁺ + SO₄²⁻

1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)

but Ba²⁺ = 1.3×10⁻² M

1.07×10⁻¹⁰  = 1.3×10⁻² M × ( SO₄²⁻)

( SO₄²⁻)  = 1.07×10⁻¹⁰  / 1.3×10⁻² = 0.823 × 10⁻⁸ M

while Ksp(CaSO4)=7.10×10−5

CaSO₄ → Ca²⁺ + SO₄²⁻

7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)

( SO₄²⁻)  = 7.10×10⁻⁵  /  2.0×10⁻² = 3.55 × 10⁻³ M

comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄

7 0
3 years ago
When the forward and reverse paths of a change occur at the same rate,
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The correct answer is option B. When the forward and reverse paths of a change occur at the same rate, <span>the system is in equilibrium specifically in dynamic equilibrium.</span> Dynamic equilibrium is the balance in a process that is continuing. It is achieved in a reaction when the forward rate of reaction and the backward rate of reaction is at the same value or equal.

8 0
3 years ago
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1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur
marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that P_{1} = 100\,kPa, V_{1} = 500\,mL and V_{2} = 1000\,mL, then the new pressure of the gas is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 500\,kPa

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that V_{1} = 3.60\,L, P_{1} = 10\,kPa and P_{2} = 25\,kPa then the new volume of the gas is:

V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)

V_{2} = 1.44\,L

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that \frac{P_{2}}{P_{1}} = 3, then the volume ratio is:

\frac{V_{1}}{V_{2}} = 3

\frac{V_{2}}{V_{1}} = \frac{1}{3}

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

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2 years ago
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Answer:

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