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3 years ago

2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) T

Chemistry

1 answer:

brilliants [131]3 years ago

6 0

Answer:

ΔHrxn = [(1) -1675.5 ( kJ/mole) + (2) 0 ( kJ/mole)] - [(1) -824.3 ( kJ/mole) + (2) 0 ( kJ/mole)]

Explanation:

ΔHrxn = 2ΔHf (Al₂O₃) - ΔHf (Fe₂O₃)

Remember that for pure elements in their standard state of temperature and pressure by definition their standard heats of formation are zero.

ΔHrxn = 2(-1675.7) - (-824.3) kJ/mol

ΔHrxn = 2527 kJ/mol

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A sample of gas has a mass of 0.560 g . Its volume is 125 mL at a temperature of 85 ∘C and a pressure of 757 mmHg .

-BARSIC- [3]

Answer:

132g/mole

Explanation:

using the formula PV=nRT should be used to solve for the number of moles (n). R is a constant which is 62.3637 L mmHG/mole K.

Inorder for your units to match you will have to convert 125ml to .125L and the temperature of 85C to K . you do that by adding 273 to the 85C and get 358K. Once you solve for n then you use that number and divide by the number of grams from the question (.560g) since molar mass is grams/moles.

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3 years ago

All of the substances on the periodic table are classified as elements except

zzz [600]

Answer: all elements in the periodic table is classified as elements

Explanation:

The structure of the table shows periodic trends. The seven rows of the table, called periods, generally have metals on the left and nonmetals on the right. The columns, called groups, contain elements with similar chemical behaviours. Six groups have accepted names as well as assigned numbers: for example, group 17 elements are the halogens; and group 18 are the noble gases

6 0

3 years ago

9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r

Mumz [18]

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

$\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O$

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

$\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\ \\ Percent\text{ }yield=0.9484\times100\% \\ \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}$

Therefore, the percent yield for the reaction is 95%.

3 0

1 year ago

How many moless is 4.91x10^22 molecules of H3PO4?

Lynna [10]

1 mol = 6.023x10^23 number of molecules (Avogadro's number)

1 : 6.023x10^23
X : 4.91x10^22

(6.023x10^23)X = 4.91x10^22

X = 4.91x10^22/6.023x10^23

X = 0.082 Moles

6 0

3 years ago

Significant Digits and Scientific Notation Part 1: Determine the number of significant digits in each number and write out the s

MA_775_DIABLO [31]

Answer:

1. 6 significant digits. 2. 2 significant digits. 3. 8 significant digits. 4. 4 significant digits. 5. 8 significant digits. 6. 7 significant digits. 7. 2 significant digits.

Explanation:

All nonzero digits are significant. All zeros before nonzeros are insignificant. All zeros after nonzeros are insignificant unless they are made for precision of some data. So 405000 may have 3 or 6 significant figures. But usually all zero figures after significant digits are significant.

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3 years ago