A chemist burns 0.8 g of iron in oxygen.

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

Summarize the acid-base behavior of the main-group metal and nonmetal oxides in water. How does oxide acidity in water change do

MaRussiya [10]

Explanation:

Metals react with oxygen to form basic oxides while they react with water to form alkaline solutions. Also, acidic oxides are oxides of nonmetals and they react with water to form acidic solutions.

Trends on the period table shows the variation of metallic character as you move across and down the periodic table. Metallic character of a element decreases across the period on the periodic table from left to right because atoms readily accept electrons in their outermost shell to form stable configurations. Metallic character increases as you move down the group in the periodic table and this is because electrons become easier to lose as the atomic radius increases (more outer shells are added), where there is decreasing attraction between the nucleus and the valence electrons.

So down the group, the acidity of oxide reaction with water decreases because the oxides are more basic down the group while across the period, the acidity of oxide increases because acidic oxides are formed as we move across the period.

4 0

3 years ago

Plants and algae that convert sunlight to food energy are called

borishaifa [10]

The correct answer is the first choice given. Plants and algaes are producers. They are called as such since they produce their own food by using the energy from the sun, CO2 and water to form glucose as their food.

8 0

3 years ago

Methanol (ch3oh), also called methyl alcohol, is the simplest alcohol. it is used as a fuel in race cars and is a potential repl

erica [24]

Answer:

%age Yield = 51.45 %

Solution:

Step 1: Convert Kg into g

68.5 Kg CO = 68500 g CO

8.60 Kg H₂ = 8600 g

Step 2: Find out Limiting reactant;

The Balance Chemical Equation is as follow;

CO + 2 H₂ → CH₃OH

According to Equation,

28 g (1 mol) CO reacts with = 4 g (2 mol) of H₂

So,

68500 g CO will react with = X g of H₂

Solving for X,

X = (68500 g × 4 g) ÷ 28 g

X = 9785 g of H₂

It shows 9785 g H₂ is required to react with 68500 g of CO but we are provided with 8600 g of H₂ which is less than required. Therefore, H₂ is provided in less amount hence, it is a Limiting reagent and will control the yield of products.

Step 3: Calculate Theoretical Yield

According to equation,

4 g (2 mol) H₂ reacts to produce = 32 g (1 mol) Methanol

So,

8600 g H₂ will produce = X g of CH₃OH

Solving for X,

X = (8600 g × 32 g) ÷ 4 g

X = 68800 g of CH₃OH

Step 4: Calculate %age Yield

%age Yield = Actual Yield ÷ Theoretical Yield × 100

Putting Values,

%age Yield = 3.54 × 10⁴ g ÷ 68800 g × 100