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Pani-rosa [81]
3 years ago
11

Please help me. The question is blow thanks.

Chemistry
1 answer:
lesantik [10]3 years ago
3 0
I'm pretty sure it's C amplitude and wavelength hope this helps
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Veronika [31]

Answer:co2

Explanation:

7 0
3 years ago
Someone help me rn! ASAP
Olenka [21]

Answer:thermal energy

Explanation:Thermal radiation is the emission of electromagnetic rays from all matter that’s greater then zero

3 0
3 years ago
A mixture of gases is analyzed and found to have the following composition in mol %: CO2 12.0 CO 6.0 CH4 27.3 H2 9.9 N2 44.8 a)
vichka [17]

Answer:

a) CO₂: <em>21,9%; </em>CO: <em>7,0%; </em>CH₄: <em>18,2%; </em>H₂: <em>0,8%; </em>N₂: <em>52,1%</em>

b) 24,09 g/mol

Explanation:

a) It is posible to obtain the composition of the gas mixture in weight% using molecular mass of each compound, thus:

12% CO₂×\frac{44,01g}{1mol} = <em>528,1 g</em>

6% CO×\frac{28,01g}{1mol} = <em>168,1 g</em>

27,3% CH₄×\frac{16,05g}{1mol} = <em>438,2 g</em>

9,9% H₂×\frac{2,02g}{1mol} = <em>20,0 g</em>

44,8% N₂×\frac{28g}{1mol} = <em>1254,4 g</em>

The total mass of the gas mixture is:

528,1g + 168,1g + 438,2g + 20,0g + 1254,4g = <em>2408,8 g</em>

Thus composition of the gas mixture in weight% is:

CO₂: \frac{528,1g}{2408,8g}×100 = <em>21,9%</em>

CO: \frac{168,1g}{2408,8g}×100 = <em>7,0%</em>

CH₄: \frac{438,2g}{2408,8g}×100 = <em>18,2%</em>

H₂: \frac{20,0g}{2408,8g}×100 = <em>0,8%</em>

N₂: \frac{1254,4g}{2408,8g}×100 = <em>52,1%</em>

b) The average molecular weight of the gas mixture is determined with mole % composition, thus:

0,12×44,01g/mol + 0,06×28,01g/mol + 0,273×16,05g/mol + 0,099×2,02g/mol + 0,448×28g/mol = <em>24,09 g/mol</em>

I hope it helps!

6 0
3 years ago
Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for
eduard

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

7 0
3 years ago
What is the volume of 0.23kg of pure water
vichka [17]
M = 0,23kg = 230g
d = 1g/cm³

V = 230g / 1g/cm³ = 230cm³ = 0,23L
7 0
3 years ago
Read 2 more answers
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