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How many square centimeters are in an area of 8.50 inch^2

Paha777 [63]

Explanation:

Given parameters:

Dimension = 8.5in²

we are to convert from in² to cm²

in² = in x in

cm² = cm x cm

1 in = 2.54cm

Using dimensional analysis;

8.5in² x $\frac{2.54cm}{1in} x \frac{2.54cm}{1in}$

= 54.84cm²

We solved this problem using dimensional analysis where we multiplied the unit with 1.

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5 0

3 years ago

What is a material which cannot be broken down or changed into another substance using chemical matters?

Alika [10]

Elements cannot be broken down in this fashion.

Compounds, homogeneous mixtures, and heterogeneous mixtures can be broken down or separated with chemical processes, or in some cases, just by manual separation.

5 0

3 years ago

How many molecules of water are in 3.8 moles of water?

Salsk061 [2.6K]

Answer:

2.29*10^24 molecules of water

Explanation:

1 molof water =6.022*10^23 molecules of water

therefore

molecules of water = 3.8*6.022*10^23

5 0

3 years ago

Read 2 more answers

Determine the mass of 5.2 x 10 power of 21 molecules of propanol C3H7OH(l), on grams.

Alja [10]

$n(\text{C}_3\text{H}_7\text{OH}) = N(\text{C}_3\text{H}_7\text{OH}) / N_A\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 5.2 \times 10^{21} / (6.02 \times 10^{23})\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 8.6 \times 10^{-3} \; \text{mol}$

where $N_A = 6.02 \times 10^{23}\; \text{mol}^{-1}$ the Avogadro's constant that relates the number of particles to their number, in the unit moles $\text{mol}$ .

The molar mass of propanol- mass per mole propanol- can be directly deduced from its molecular formula with reference to a modern periodic table.

$M(\text{C}_3\text{H}_7\text{OH}) = \underbrace{3 \times 12.01}_{\text{carbon}} + \underbrace{8 \times 1.008}_{\text{hydrogen}} + \underbrace{1\times 16.00}_{\text{oxygen}} = 60.09 \; \text{g} \cdot \text{mol}^{-1}$

$8.6 \times 10^{-3} \; \text{mol}$ of propanol molecules would thus have a mass of $8.6 \times 10^{-3} \; \text{mol} \times 60.09 \; \text{g} \cdot \text{mol}^{-1} = 0.52 \; \text{g}$

5 0

4 years ago

Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance

ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0

3 years ago