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3 years ago

Given the number of a substance , how do you solve for the number of a moles of a substance

1 answer:

5 0

Given the number of a substance, we can solve the number of moles by using a conversion factor that would relate the number of a substance to the number of moles. In any case, Avogadro's number would be used. It represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. For example, we are given 1.23 x 10^24 atoms of a substance converting it to moles we do as follows:

1.23 x 10^24 atoms ( 1 mol / 6.022x10^23 atoms ) = 2.04 moles

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a 160 milligram sample of a radioactive isotope decays to 10 kilograms in 12 years. what is the half life of this element

Shkiper50 [21]

Answer:

3 years

Explanation:

Given data:

Initial amount of sample = 160 Kg

Amount left after 12 years = 10 Kg

Half life = ?

Solution:

at time zero = 160 Kg

1st half life = 160/2 = 80 kg

2nd half life = 80/2 = 40 kg

3rd half life = 40 / 2 = 20 kg

4th half life = 20 / 2 = 10 kg

Half life:

HL = elapsed time / half life

12 years / 4 = 3 years

8 0

3 years ago

a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v

Rashid [163]

Concentration of NaOH = 0.148 molar, M

Concentration of H3PO4 = 0.172 molar, M

Concentration x Volume will give the number of moles of solute in that volume. C*V = moles

Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:

NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:

(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M

====

The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:

(0.00172 moles H3PO4)/(0.010 liters H3PO4)

Concentration of H3PO4 = 0.172 molar, M

5 0

1 year ago

Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 °C? (HINT: the act

algol13

Answer:

D) 2.3 x 10⁻¹ s⁻¹

Explanation:

The rate constant is related to the activation energy through the formula:

k= Ae^(-Eₐ /RT)

where A is the collision factor, Eₐ the activation energy, R is the gas constant ( 8.314 J/Kmol ) , and T is the temperature (K)

So a plot of lnk versus 1/T ( Arrehenius plot ) gives us a straight line with slope equal -Eₐ/R and intercept lnA

lnk = -(Eₐ/T)(1/T) + lnA

which has the form y= mx + b

In this problem, we can use the data provided to:

a) Using a calculator determine the slope and intercept and then calculate the value of rate constant at 320 ºC, or

b) Plot the data and determine the equation of the best line , and answer the question for k @ 320 ºC by reading the value from the plot.

Once you do the plot, the resulting equation is:

y = - 19 x 10³ x + 30,582 ( R² = 0.999 )

So for T = 320 + 273 K = 593 K

Y = 19 x 10³ X + 30.58

So for T = (320 + 273)K = 593 K

Y = -19 x 10³ ( 1/593) + 30.58 = -32.04 +30.58 = - 1.46

and then since

y = lnk ⇒ e^y = k

k= e^-1.46 = 2.3 x 10⁻¹ s⁻¹

Note: there is an error of transcription in the value for T = 472.1 ( 1/T = 2.118 x 10⁻³ and not 2.228 x 10⁻³). You can recognize this mistake if you plot the data and notice it produces an outlier.

5 0

3 years ago

A reaction has a rate constant of 1.15 x 10^−2 /s at 400K and 0.685 /s at 450K.

n200080 [17]

Answer:

a) the activation barrier = 122.3 kJ/mol

b) The rate constant at 425 K = 0.1001 /s

Explanation:

Step 1: Data given

Rate constant k1 = 1.15 * 10^−2 /s at 400K (= T1)

Rate constant k2 = 0.685 /s at 450K (=T2)

Step 2: Determine the activation barrier for the reaction.

To determine the activation energy we will use the two-point Arrhenius equation:

ln(k₂/k₁) = (Ea/R)((1/T1) - (1/T2))

⇒ with Ea = the activating energy

⇒ with R = the gas constant = 8.314 J/mol* K

⇒ with k1 = rate constant 1 = 1.15 *10^-2 /s

⇒ with T1 = Temperature 1 = 400 K

⇒ with k2 = rate constant 2 = 0.685/s

⇒ with T2 = temperature 2 = 450 K

= - (Ea/R)(T₁ - T₂)/T₁T₂

Ea = (R*ln (k2/k1)) / ((1/T1)- (1/T2))

Ea = (8.314* ln(0.685/0.0115)) / ((1/400) - (1/450))

Ea = 122327.6 = 122.3 kJ/mol

B) What is the value of the rate constant at 425 K

For rate constant at 425 K.

Substitute the value of activation energy as 122327.6 J/mol, initial temperature as 400 K, final temperature as 425 K, rate constant at 400 K

1/T1 - 1/ T3 = 1/400 - 1 /425 = 1.47*10^-4

⇒ with T1 = the initial temperature = 400 K

⇒ with k1 = the rate constant at 400 K = 1.15 * 10^-2 /s

⇒ with T3 = the nex temperature = 425 K

⇒ with k3 = the rate constant at 425 K

ln(k3/k1) = Ea/R * ((1/T1)- (1/ T3))

⇒ with k3 = the rate constant at 425 K

⇒ with T3 = 425 K

k3/k1 = e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = k1* e^(Ea/R * ((1/T1)- (1/ T3)))

k3 = 0.0115 * e^(122327.6/8.314 * (1.4710^-4))

k3 = 0.0115* e^2.1643

k3 = 0.1001 /s

4 0

3 years ago

what biome has many nocturnal animals that burrow underground during the day that are active at night

hram777 [196]

Answer:

it could be maybe a raccoon or a possum

Explanation:

4 0

3 years ago