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3 years ago

What is the MOST accurate explanation of the vibration of the windows?

Chemistry

1 answer:

Bumek [7]3 years ago

3 0

Answer:

Resonance

Explanation:

Like music or wind frequencies bounce off each other from windows creating a vibration

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I need help ' please

Shalnov [3]

Answer:

s orbitals - spherical shape

p orbitals - dumbbell shape

d orbitals- dxy, dyz , dzx -double dumbbell

- dx²-y²- double dumbbell (along axis)

-dz² -dumbbell with electron cloud

( along the axis)

therefore here the answer is 2s and 3s

3 0

3 years ago

How many molecules of HCl would react with 2 moles of Al?<br><br> 2Al + 6HCl → 2AlCl3 + 3H2

Yuki888 [10]

Answer:

3.612 × 10^24 molecules

Explanation:

The balanced chemical equation in this question is given as;

2Al + 6HCl → 2AlCl3 + 3H2

Based on the equation, 2 moles of Aluminum (Al) is needed to react with 6 moles of HCl

Since 1 mole of a substance is equal to 6.02 × 10^23 molecules;

This means that, according to the question, 6 moles of HCl needed to react with 2 moles of Al will contain;

6 × 6.02 × 10^23 molecules

= 36.12 × 10^23

= 3.612 × 10^24 molecules

In a nutshell, 3.612 × 10^24 molecules of HCl would react with 2 moles of Al.

5 0

3 years ago

Read 2 more answers

Why doesn't your consist of 365 days and A day of 24 hours

LenKa [72]

Hi what do you mean by this? i dont understand at all pls explain :)

8 0

3 years ago

Read 2 more answers

A natural water with a flow of 3800 m3/d is to be treated with an alum dose of 60 mg/L. Determine the chemical feed rate for the

svet-max [94.6K]

Explanation:

First, we will calculate the feed rate of alum as follows.

$\frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}$

= 228000 g/day

Converting this amount into g/min as follows.

$\frac{228000 g}{1 day} \times \frac{1 day}{1440 min}$

= 158 g/min

Now, the chemical equation will be as follows.

$Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O$

$\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}$

= 0.151 mmol $mmol SO^{2-}_{4}/L$

$\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}$

= 15.15 mg $CaCO_{3}$ /L

For precipitate:

$Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O$

$\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}$

= 7.88 $Al(OH)_{3}/L$

$\frac{7.88 mg Al(OH)_{3}}{1 L} \times \frac{3800 m^{3}}{1 day} \times \frac{1000 L}{1 m^{3}} \times \frac{1 kg}{10^{6} mg}$

= 29.9 $Al(OH)_{3}/day$

3 0

4 years ago

What is the percentage of propane in natural gas??

love history [14]

It’s a gas which causes pollution which make people cough and die

4 0

4 years ago