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3 years ago

When can the park be reopen to people and animals. Explain calculations and answers. Explain the concept.

Chemistry

2 answers:

julsineya [31]3 years ago

5 0

Answer:

in 23 years

Explanation:

Minchanka [31]3 years ago

5 0

The answer would be the park can reopen to people and animals in 23 years.

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What are the characteristics of a continental drift?

Eduardwww [97]

Continental drift was a theory that explained how continents shift position on Earth's surface. Set forth in 1912 by Alfred Wegener, a geophysicist and meteorologist, continental drift also explained why look-alike animal and plant fossils, and similar rock formations, are found on different continents.

7 0

3 years ago

Please help me im going out of my way and giving you extra points I really need some assistance

Hitman42 [59]

Answer:

x=2.8moles

Explanation:

first step balance the chemical equation

C2H6+O2-->CO2+2H2O

use more ratio to find the moles of water

1mole of C2H6= 2 moles of H2O

1.4mole of C2H6=?x

cross multiply

x=2.8 moles of H2O

4 0

3 years ago

2.1 g of a hydrocarbon fuel is burned in a calorimeter that contains 280 grams of water initially at 25.00◦C. After the combusti

olga2289 [7]

Answer:

$\large \boxed{\text{933 J}}$

Explanation:

There are three heat transfers involved.

heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0

q₁ + q₂ + q₃ = 0

m₁ΔH + m₂C₂ΔT + C_calΔT = 0

Data:

m₁ = 2.1 g

m₂ = 280 g

Ti = 25.00 °C

T_f = 26.55 °C

Ccal = 92.3 J·°C⁻¹

Calculations:

Let's calculate the heats separately.

1. q₁

q₁ = 2.1 g × ΔH = 2.1ΔH g

2. q₂

ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C

q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J

3. q₃

q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J

4. ΔH

$\begin{array}{rcl}\text{2.1$\Delta$H g +1816 J +143.1 J} & = & 0\\\text{2.1$\Delta$H g +1959 J} & = & 0\\\text{2.1$\Delta$H g}& = & \text{-1959 J}\\\Delta H & = & \dfrac{\text{-1959 J}}{\text{2.1 g}}\\\\& = & \textbf{-933 J/g}\\\end{array}\\\text{The combustion releases $\large \boxed{\textbf{933 J}}$ per gram of fuel burned.}$

7 0

3 years ago

A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve

mixas84 [53]

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

PV = nRT

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

= $\frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}$

= 0.0052 mol

Also, No. of moles = $\frac{mass}{\text{molar mass}}$

0.0052 mol = $\frac{mass}{2 g/mol}$

mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at $25^{o}C$

$P_{\text{water vapor}}$ = 24 mm Hg

= $\frac{24}{760}$ atm

= 0.03158 atm

Now, P = $\frac{756}{760} - 0.03158$

= 0.963 atm

Hence, n = $\frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}$

= 0.0056 mol

So, mass of $H_{2}$ = 0.0056 mol × 2

= 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$

= $\frac{0.01013 g}{0.0104 g} \times 100$

= 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

3 0

3 years ago

What is the specific heat capacity of an unknown metal if 75.00 g of the metal absorbs 418.6J of heat and the temperature rises

EleoNora [17]

Answer:

The specific heat capacity of the unknown metal is 0.223 $\frac{J}{g*C}$

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

Q= 418.6 J
c= ?
m= 75 g
ΔT= 25 C

Replacing:

418.6 J= c* 75 g* 25 C

Solving:

$c=\frac{418.6 J}{75 g*25 C}$

c= 0.223 $\frac{J}{g*C}$

The specific heat capacity of the unknown metal is 0.223 $\frac{J}{g*C}$

3 0

2 years ago