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Alika [10]
3 years ago
8

Find the volume. * 1) 6 in 11 in Volume =

Chemistry
1 answer:
posledela3 years ago
7 0

Known :

r = 6 in

h = 11 in

Asked :

Volume = ...?

Answer :

V = πr²h

= 3.14 × 6² × 11

= 3.14 × 36 × 11

= 3.14 × 396

= <u>1</u><u>,</u><u>2</u><u>4</u><u>3</u><u>.</u><u>4</u><u>4</u><u> </u><u>in³</u>

So, the volume is 1,243.44 in³ (2nd choice)

<em>Hope </em><em>it </em><em>helpful </em><em>and </em><em>useful </em><em>:</em><em>)</em>

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A sample of carbon monoxide gas is initially in a 5858 mL container. The gas is then moved to 3.29 L container at a temperature
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Answer:

d i took the test

Explanation:

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A sound wave travels from air that is at 24°C into air that is 15°C. The sound wave will _____.
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The sound wave will Slow Down
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Which postulate of Dalton's atomic theory was later proven wrong?
ruslelena [56]

Answer:

Option c and d

Explanation:

John Dalton. In 1808, John Dalton proposed a theory known as Dalton’s Atomic Theory. The theory was published in a paper titled “A New Chemical Philosophy”. This theory was new to that era

The 5 postulates of Daltons' atomic theory are:

1. All the matters are made of atoms.

2. Atoms of different elements combine to form compounds

3. Compounds contain atoms in small whole-number ratios

4. Atoms can neither be created nor destroyed . (This was later proven wrong )

5. All atoms of an element are identical and have the same properties (This was later proven wrong as atoms of same element may be different in case of elements having isotopes )

Therefore, options c and d are the answer.

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The pen on seismograph swings freely.
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3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
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