If a solution at pH 5 undergoes a 1000-fold increase in [OH-], what is the resulting pH?

Thallium-207 decays exponentially with a half life of 4.5 minutes. if the initial amount of the isotope was 28 grams, how many g

Anarel [2728]

An exponential decay law is generally expressed as: A = Ao * e ^ (-kt) => A/Ao = e^(-kt) Half-life time => A/Ao = 1/2, and t = 4.5 min => 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154. Now substituting k, Ao = 28g, and t = 7 min to determine the remaining grams of Thallium-207 gives: A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g. Final answer is 9.5 g.

7 0

20 days ago

"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92

lorasvet [2668]

Answer: The percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

From the information provided:

Let the fractional abundance for $_{14}^{28}\textrm{Si}$ isotope be 'x'

For $_{14}^{28}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 27.9769 amu

Percentage abundance of $_{14}^{28}\textrm{Si}$ isotope = 92.22 %

Fractional abundance for $_{14}^{28}\textrm{Si}$ isotope = 0.9222

For $_{14}^{29}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 28.9764 amu

Percentage abundance for $_{14}^{28}\textrm{Si}$ isotope = 4.68%

Fractional abundance of $_{14}^{28}\textrm{Si}$ isotope = 0.0468

For $_{14}^{30}\textrm{Si}$ isotope:

Mass of $_{14}^{30}\textrm{Si}$ isotope = 29.9737 amu

Fractional abundance for $_{14}^{30}\textrm{Si}$ isotope = x

The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

$28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309$

To convert this fractional abundance into a percentage, multiply by 100:

$\Rightarrow 0.0309\times 100=3.09\%$

This shows that the percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

6 0

1 month ago

A 20.0–milliliter sample of 0.200–molar K2CO3 solution is added to 30.0 milliliters of 0.400–molar Ba(NO3)2 solution. Barium c

KiRa [2853]

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

O sea,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Dado que:

Para $K_2CO_3$ :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de $K_2CO_3$ son:

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles de $K_2CO_3$ = 0.004 moles

Para $Ba(NO_3)_2$ :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de $Ba(NO_3)_2$ son:

$Moles=0.400 \times {30.0\times 10^{-3}}\ moles$

Moles de $Ba(NO_3)_2$ = 0.012 moles

Según la reacción:

$Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3$

1 mol de $Ba(NO_3)_2$ reacciona con 1 mol de $K_2CO_3$

Por lo tanto,

0.012 mol de $Ba(NO_3)_2$ reacciona con 0.012 mol de $K_2CO_3$

Moles disponibles de $K_2CO_3$ = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces $K_2CO_3$ es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de $K_2CO_3$ reacciona con 1 mol de $Ba(NO_3)_2$ y produce 1 mol de $BaCO_3$

0.004 mol de $K_2CO_3$ reacciona con 0.004 mol de $Ba(NO_3)_2$ y genera 0.004 mol de $BaCO_3$

Los moles restantes de $Ba(NO_3)_2$ son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, $Ba^{2+}$ , después de la reacción es:

$Molarity=\frac{0.008}{0.050}\ M = 0.16\ M$

3 0

1 month ago

The chemical formula for cesium oxide is Cs2O. What is the charge of cesium? +1 +2 –1 –2

lions [2782]

The charge on cesium is +1

5 0

1 month ago

Read 2 more answers

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

lorasvet [2668]

Answer:

Can you rank the following chemical substances in order of their absolute entropies (So) from lowest (1) to highest (5) at a temperature of 298 K?

a. Al (s)

b. H2O (l)

c. HCN (g)

d. CH3COOH (l)

e. C2H6 (g)

Explanation:

Entropy quantifies the level of disorder within a system.

In solids, the entropy is significantly lower compared to liquids and gases.

The typical order of entropy is:

solids < liquids < gases

In the substances listed, liquid water notably exhibits strong intermolecular hydrogen bonding.

This results in water having comparatively lower entropy.

Next in line is acetic acid.

Among the gaseous components, ethane has higher entropy than HCN due to its weaker intermolecular interactions.

HCN involves some hydrogen bonding.

Thus, the order of entropy is:

Al(s) < CH3COOH (l) < H2O(l) < HCN(g) < C2H6(g)

7 0

1 month ago