When the volume and number of particles of a gas are constant which of the following is also constant

The radioisotope that has the longest half-life is the best to use in powering planet and space exploration vehicles because the

Diano4ka-milaya [45]

Answer:

U-238

Explanation:

For a given radioisotope, half life can be defined as the time taken for the isotope to decay into one-half of its original amount. Mathematically, this is expressed as:

$t_{1/2} = \frac{0.693}{k}$

where k = rate constant for the radioactive decay process

Greater the t1/2 longer will be its stability.

Based on the given data, U-238 has the largest half life and therefore will be best suited for applications mentioned.

4 0

3 years ago

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Kernels of corn are heated and become popcorn. Why is this a chemical change?

goldfiish [28.3K]

Answer:

When popcorn is popped, liquid inside the kernel is changed to steam. Pressure from the steam builds up inside the kernel. When the pressure reached a critical stage the kernel pops turning itself inside out. This is a physical change.

Explanation:

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3 years ago

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What are the concentrations of A , A, B , B, and C C at equilibrium if, at the beginning of the reaction, their concentrations a

Elenna [48]

The question is incomplete, here is the complete question:

A reaction

$A+B\rightleftharpoons C$

has a standard free-energy change of -4.88 kJ/mol at 25°C

What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30 M, 0.40 M and 0 M respectively?

Answer: The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

Explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_c$

where,

$\Delta G^o$ = Standard Gibbs free energy = -4.88 kJ/mol = -4880 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_c$ = equilibrium constant of the reaction

Putting values in above equation, we get:

$-4880J/mol=-(8.3145J/Kmol)\times 298K\times \ln K_c\\\\K_c=7.17$

We are given:

Initial concentration of A = 0.30 M

Initial concentration of B = 0.40 M

Initial concentration of C = 0 M

The chemical reaction follows:

$A+B\rightleftharpoons C$

Initial: 0.30 0.40 0

At eqllm: 0.30-x 0.40-x x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[C]}{[A][B]}$

We are given:

$K_c=7.17$

Putting values in above equation, we get:

$7.17=\frac{x}{(0.30-x)\times (0.40-x)}\\\\x=0.183,0.657$

Neglecting the value of x = 0.657, because change cannot be greater than the initial concentration

So, equilibrium concentration of A = $(0.30-x)=(0.30-0.183)=0.117M$

Equilibrium concentration of B = $(0.40-x)=(0.40-0.183)=0.217M$

Equilibrium concentration of C = $x=0.183M$

Hence, the equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

5 0

3 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

3 years ago

Sodium bicarbonate (NaHCO3) is commercially known as baking soda. How many grams and atoms of oxygen are there in 15 grams of so

ella [17]

Answer:

3 atoms of O

8.57g

Explanation:

The problem here is to find the mass of oxygen in the compound and the number of atoms of oxygen.

The formula of the compound is:

NaHCO₃;

Here we have:

1 atom of Na

1 atom of H

1 atom of C

3 atoms of O

So there are 3 atoms of O

To find the mass of oxygen , we use the molar mass of oxygen;

Molar mass of oxygen = 16g/mol

Mass of oxygen = $\frac{3(16)}{84.01}$ x 15 = 8.57g

4 0

3 years ago