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3 years ago

Change is still the same 16. A substance that undergoes a substance after the change.

Chemistry

1 answer:

V125BC [204]3 years ago

3 0

Answer:

When substances undergo chemical change, the mass of the reactants will be the same as the mass of the products. Mass is converted when substances undergo chemical change but is not lost. Mass cannot be created or destroyed.

Explanation:

i hope this helps :)

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When potassium chlorate decomposes to potassium chloride and oxygen gas is it a physical or chemical change?

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It is an example of a Chemical change.

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Determine the partial pressure and number of moles of each gas in a 15.75-L vessel at 30.0 C containing a mixture of xenon and n

lawyer [7]

Answer:

The Partial pressure of Xe and Ne will be 4.95 atm and 1.55 atm. The number of moles of Xe and Ne will be 3.13 and 0.981

Explanation:

Let the total pressure of the vessel= 6.5 atm and mole fraction of Xenon= 0.761

As we know,

$\chi_{Ne} + \chi_{Xe} = 1\\\chi_{Ne}= 1- 0.761\\\chi_{Ne}= 0.239$

According to Dalton's Law of partial pressure-

$P_i=\chi_i\times P_{total}$

Where,

$P_i=$ The pressure of the gas component in the mixture

$\chi_i=$ Mole fraction of that gas component

$P_t=$ The total pressure of the mixture

$P_{Xe}=(0.761)\times(6.5)\\P_{Xe}= 4.95 atm\\\\\\P_{Ne}=(0.239)\imes (6.5)\\P_{Ne}= 1.55 atm$

<u>Calculation: </u>

To calculate the number of moles,

PV=nRT

$n=\frac{PV}{RT}$

$n_{Xe}= \frac{4.95\times 15.75}{0.0821\times303 }\\ n_{Xe}= \frac{77.96}{24.87} \\n_{Xe}= 3.13\,mole \\\\\\n_{Ne}= \frac{1.55\times 15.75}{0.0821\times303 }\\\\n_{Ne}=\frac{24.41}{24.87}\\ n_{Ne}=0.981 \,mole$

Learn more about Dalton's Law of partial pressure here;

brainly.com/question/14119417

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1 year ago

Energy and Enthalpy Changes, Heat and Work -- Monatomic Ideal Gas dynamically generated plot 2.00-mol of a monatomic ideal gas g

larisa86 [58]

(a) The heat generated in the process is 28 kJ.

(b) The work done in the process is determined as -28 kJ.

<h3>Heat of the isothermal compression </h3>

The heat generated in the process is negative done in the process.

W = -PΔV

W = -P(V₂ - V₁)

W = -P(VB - VA)

W = -11(7 - 12.5)

W = 60.5 L.atm = 60.5 x 101.325 J/L.atm = 6,130.16 J

W = -25(20.5 - 7)

W = -337.5 L.atm = -34,197.18 J

Total work , w = -34,197.18 J + 6,130.16 J = -28 kJ

q = - w

q = 28 kJ

<h3>Change in internal energy</h3>

ΔE = q + w

ΔE = 28 kJ - 28 kJ = 0

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