According to the cup wall diagrams, why does the double wall vacuum

Iron is extracted from iron oxide in the Blast Furnace: Fe 2 O 3 + 3 CO → 2 Fe + 3 CO 2

arsen [322]

a. mass of iron = 69.92 g

b. percent yield = 93%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

a.

Reaction

Fe₂O₃+3CO⇒2Fe+3CO₂

MW Fe₂O₃ : 159.69 g/mol

mol Fe₂O₃

$\tt \dfrac{100}{159,69}=0.626$

mol Fe₂O₃ : mol Fe = 1 : 2

mol Fe :

$\tt \dfrac{2}{1}\times 0.626=1.252$

mass of Fe(Ar=55.845 g/mol) :

$\tt 1.252\times 55.845=69.92~g$

b.

actual yield = 65 g

theoretical yield = 69.92 g

percent yield :

$\tt =\dfrac{65}{69.92}=0.93=93\%$

8 0

3 years ago

55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?

ladessa [460]

Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}$ ,

where

$V_2$ is the final volume,
$V_1$ is the initial volume,
$T_2$ is the final temperature in degrees Kelvins.
$T_1$ is the initial temperature in degrees Kelvins.

Convert the temperatures to degrees Kelvins:

$T_1 = \rm 25^{\circ}C = (25 + 273.15)\; K = 298.15\; K$ .

$T_2 = \rm 35^{\circ}C = (35 + 273.15)\; K = 308.15\; K$ .

Apply Charles's Law to find the new volume of this gas:

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 55\;L \times \frac{308.15\; K}{298.15\; K} = 56.8\; L$ .

8 0

3 years ago

What Resources do organisms compete over?

Alex787 [66]

Siekejekekekskkensndjdjd

7 0

3 years ago

0.40 L of HNO3 is titrated to equivalence using 0.18 L of 0.1 M NaOH. What is the concentration of the HNO3 ?

e-lub [12.9K]

This problem is providing us with the volume of nitric acid that is titrated with 0.18 L of 0.1-M sodium hydroxide and asks for the concentration of the acid. At the end, the result turns out to be 0.045M, according to the following.

<h3>Acid-base titrations:</h3>

In chemistry, acid-base titrations allow us to quantify the volume or concentration of an acid or base via the following equation:

$M_AV_A=M_BV_B$

Where the subscript A stands for the acid and B for the base; which means one can calculate any of the variables there by knowing the other three. This equation is based on the balanced neutralization chemical equation, which takes place between the acid and the base.