If a radioisotope has a half-life of 4 years, how much remains of a starting amount of 40 grams after 16 years have passed?

Why does permeable rock weather more than less permeable rock

Dafna11 [192]

<span>Permeable means that a material is full of tiny, connected airspaces that allows water to seep through it.</span>

5 0

4 years ago

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A GFCI is an example of what kind of electrical safety device?

vichka [17]

Answer:

Circuit breaker.

Explanation:

3 0

3 years ago

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If the reaction of 23.3 grams of Cr2O3 produces 5.35 grams<br> of Al2O3, what is the percent yield?

Marat540 [252]

Answer:

34.2%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Cr₂O₃ + 2Al —> Al₂O₃ + 2Cr

Next, we shall determine the mass of Cr₂O₃ that reacted and the mass of Al₂O₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Cr₂O₃ = (52×2) + (3×16)

= 104 + 48

= 152 g/mol

Mass of Cr₂O₃ from the balanced equation = 152 × 1 = 152 g

Molar mass of Al₂O₃ = (27×2) + (16×3)

= 54 + 48

= 102 g/mol

Mass of Al₂O₃ from the balanced equation = 1 × 102 = 102 g

SUMMARY:

From the balanced equation above,

152 g of Cr₂O₃ reacted to produce 102 g of Al₂O₃.

Next, we shall determine the theoretical yield of Al₂O₃. This can be obtained as follow:

From the balanced equation above,

152 g of Cr₂O₃ reacted to produce 102 g of Al₂O₃.

Therefore, 23.3 g of Cr₂O₃ will react to produce = (23.3 × 102)/152 = 15.64 g of Al₂O₃.

Thus, the theoretical yield of Al₂O₃ is 15.64 g.

Finally, we shall determine the percentage yield of Al₂O₃. This can be obtained as follow:

Actual yield of Al₂O₃ = 5.35 g

Theoretical yield of Al₂O₃ = 15.64 g.

Percentage yield of Al₂O₃ =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.35 / 15.64 × 100

Percentage yield of Al₂O₃ = 34.2%

3 0

3 years ago

Calcium carbonate decomposes at high temperatures to give calcium oxide and carbon dioxide as shown below. CaCO3(s) CaO(s) + CO2

romanna [79]

Answer:

$\boxed{\text{10.0 g}}$

Explanation:

The balanced equation is

CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 1.16

Calculate Kc

T= 800 °C = 1073 K; Δn= 1

$K_{\text{p}} = K_{\text{c}}{RT}^{\Delta n}\\\\1.1 = K_{c}{(RT)}^{1}\\\\K_{c} = \dfrac{1.1}{RT} = \dfrac{1.1}{8.314\times 1073}= \dfrac{1.1}{8921} = 1.233 \times 10^{-4}$

Equilibrium concentration of CO₂

$K_{c} = [\text{CO}_{2}] = 1.233 \times 10^{-4}$

Moles of CO₂ formed

$n= \text{5.00 L } \times \dfrac{\text{1.233 $\times$ 10$^{-4}$ \text{ mol} }}{\text{1 L}} = 6.155 \times 10^{-4}\text{ mol}$

Moles of CaCO₃ used up

Moles of CaCO₃ used up = moles of CO₂ formed = 6.155 × 10⁻⁴ mol

Mass of CaCO₃ used up

$m = 6.155 \times 10^{-4}\text{ mol } \times \dfrac{\text{100.09 g}}{\text{1 mol}} = \text{0.0616 g}$

Moles of CaO formed

Moles of CaO formed = moles of CO₂ formed = 6.155 × 10⁻⁴ mol

Mass of CaO formed

$m = 6.155 \times 10^{-4}\text{ mol } \times \dfrac{\text{56.08 g}}{\text{1 mol}} = \text{0.0345 g}$

Mass of solid at equilibrium

m = 10.0 g – 0.0616 g + 0.0345 g = 10.0 g

$\text{The mass of solid remaining at equilibrium is } \boxed{\textbf{10.0 g}}$

6 0

3 years ago

On a hot sunny day, you get out of the swimming pool and sit in a metal chair, which is very hot. Would you predict that the spe

Assoli18 [71]

Answer:

The specific heat capacity of the metal is lower than that of the water as explained as follows

Explanation:

The heat capacity is a measure of propensity of a given object or matter to undergo an increase or decrease in its temperature when subject a given amount of heat, in other words, for a given mass of the substance, it is the amount of heat required to be added to cause it to have a unit temperature change

The specific heat capacity of a material or substance is the amount of heat required to be supplied to raise the temperature of a unit mass of that by one degree Celsius. It is the ratio of the heat capacity of a specimen of a substance divided by the mass of the specimen

Both the water of the swimming pool, the swimmer and the metal chair are all exposed to the same amount of heat from the sun, however the heat required to raise the water temperture by one degree Celsius is higher than heat required to raise the temperature of the metal chair by one degree hence the chairs temperature is higher than the water temperature as shown in the following equation

ΔH = m₁ × C₁ × ΔT₁ = m₂ × C₂ × ΔT₂

Where ΔH = Heat or energy supplied from the sun

m₁ = mass of water in the swimming pool

C₁ = Specific heat capacity of the water in the swimming pool

ΔT₁ = Change in temperature of the swimming pool and

m₂ = mass of metal chair

C₂ = Specific heat capacity of the metal chair material

ΔT₂ = Change in temperature of the metal chair

Where ΔH is the same for both the chair and swimming pool with the chair being hotter than the swimming pool with an asumed temperature change of 1.5 times that of the swimming pool and assuming the same mass of both the swimming pool and the chair are measured we have

m₁ × C₁ × 1.5ΔT₂ = m₁ × C₂ × ΔT₂ then cancelling like terms we have

1.5C₁ = C₂

Hence the heat capacity of the swimming pool water is 1.5 times that of the metal chair

3 0

3 years ago