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Delicious77 [7]
2 years ago
8

If a wave produced by a fireworks explosion is that light wave or sound wave

Chemistry
2 answers:
gregori [183]2 years ago
8 0

Answer:

The wave produced by a fireworks "explosion" is a light wave.

Explanation:

Vesnalui [34]2 years ago
5 0

Answer:

sound wave

Explanation:

The answer is sound waves because you the sound and it makes a echo.

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How is the number of electrons in an element determined?
elixir [45]

The number of protons, neutrons, and electrons in an atom can be determined from a set of simple rules. The number of protons in the nucleus of the atom is equal to the atomic number (Z). The number of electrons in a neutral atom is equal to the number of protons.

3 0
2 years ago
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A sample of seawater weighs 250 g and has a volume of 173 ml. What is the density of seawater?
kari74 [83]

Answer:

<h2>The answer is 1.45 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass of seawater = 250 g

volume = 173 mL

It's density is

density =  \frac{250}{173}  \\  = 1.4450867052...

We have the final answer as

<h3>1.45 g/mL</h3>

Hope this helps you

8 0
2 years ago
Help please. ASAP.
Flauer [41]

Answer:

I believe- If there is more metal for the magnet to reach for, then the strength will grow but it may be harder to pick up off of a surface depending on the weight of the nail. If it is a light nail, it would be more efficient for a quick result in having the magnet to pick it up

Explanation:

4 0
2 years ago
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What volume of 0.585 m ca(oh)2 would be needed to neutralize 15.8 l of 1.51 m hcl?
SpyIntel [72]
When the balanced reaction equation is:

2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)

from the balanced equation, we can get the molar ratio between HCl & Ca(OH)2

2:1

∴ the volume of Ca(OH)2 = 15.8 L HCl * 1.51 m HCl * (1mol Ca(OH)2/ 2mol HCl) *                                           (1L ca(OH)2/0.585 mol Ca(OH)2 

                                          = 20.4 L
8 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
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