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Nostrana [21]
3 years ago
15

Sound waves travel faster than light waves. A. True B. False

Chemistry
2 answers:
Semmy [17]3 years ago
7 0
Answer: B, it is false
NikAS [45]3 years ago
4 0

Answer:

B

Explanation:

Light travels faster than sound.

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Calculate the number of moles and molecules of acetic acid in 22gm of it​
aivan3 [116]

Answer:

0.366 moles to the nearest thousandth.

Explanation:

The molar mass of acetic acid CH3COOH = 2*12 + 4(1.008) + 2*16

= 60.03 g so the number of moles in 22 g

= 0.366.

3 0
3 years ago
Why do we need to use moles when we try to determine amounts of reactants and products in a reaction?.
ICE Princess25 [194]

\huge\fbox{Answer ☘}

<em>Chemists use the mole unit to represent 6.022 × 10 23 things, whether the things are atoms of elements or molecules of compounds. This number, called Avogadro's number, is important because this number of atoms or molecules has the same mass in grams as one atom or molecule has in atomic mass units. </em>

hope helpful~

8 0
2 years ago
Why should scientists control most possible variables in their experiments?
irina [24]
Scientists should control most possible variables in experiments to get the most valid and correct data. If many variables are included in experiments it is more difficult to interpret what is causing a different outcome.
8 0
3 years ago
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
3 years ago
How many atoms are in 1.50 moles of fluorine gas
evablogger [386]
1.8x10^24 atoms is How many atoms are in 1.50 moles of fluorine gas
8 0
3 years ago
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