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Margarita [4]
3 years ago
14

Periodic Table Escape Enter the correct 4 digit code (no spaces) *

Chemistry
1 answer:
11111nata11111 [884]3 years ago
6 0

Answer:

What are you asking???

Explanation:

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A piece of metal that has a density of 5.2 g/cm3 and a mass of 100 g was placed in a full jar of water. How many mL of water was
Klio2033 [76]

Answer:

100g / (5.2g/cm3)

= 100g / (5.2g / 1cm3)

= 100g x 1cm3 / 5.2 g

= 19.2 cm3

Since 1 cm3 = 1 ml, your answer is 19.2 ml.

5 0
3 years ago
When an atom undergoes beta decay, wich of the following is true?
tigry1 [53]
<span> The atomic number increases by one and the element becomes a different element. </span>
3 0
3 years ago
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -&gt; 2Al2O3 In a certain experiment, 4.6g Al was react
stiv31 [10]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of Al produce 2 moles of Al₂O₃</em>

<em />

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

<h3>78.2% </h3>
8 0
3 years ago
A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres
tatiyna

Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 m^{3}

Mass of the gas = 1.15 gm = 0.00115 kg

Temperature = 25 ° c = 298 K

Gas constant for Argon R = 208.13 \frac{J}{kg k}

From ideal gas equation P V = m RT

⇒ P = \frac{m R T}{V}

Put all the values in above formula we get

⇒ P = \frac{0.00115}{0.001} × 208.13 × 298

⇒ P = 71.326 K pa

Therefore, the partial pressure of argon in the flask = 71.326 K pa

4 0
3 years ago
Look at the image below:
PIT_PIT [208]

Answer:

The figure is a molecule and a compound

Explanation:

Because it have 2 cluster molecules

7 0
3 years ago
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