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Radda [10]
3 years ago
9

Select the substance that is thought to be partially responsible for depleting the concentration of ozone in the stratosphere. s

elect the substance that is thought to be partially responsible for depleting the concentration of ozone in the stratosphere. o2 cfcl3 n2 he co2
Chemistry
1 answer:
gogolik [260]3 years ago
6 0

Answer:

cfcl3

Explanation:

When the cfcl3 in the atmosphere is hit by UV rays from the sun, it decomposes to Cl atoms. These Cl atoms then react with the ozone producing O2 and ClO. The ClO bombard with O atoms splitting the ClO up and the free Cl atom again bonds to another O3 molecule hence reducing it. One ClO can reduce several O3 molecules (up to 100,000) creating an ozone hole.

2Cl + O3 → 2ClO + 2O2

2ClO + 2O → 2Cl + 2O2

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Is the prime meridian a longitudinal or latitudinal line?
Nesterboy [21]

Answer:

longitudinal

Explanation:

A prime meridian is the meridian (a line of longitude) in a geographic coordinate system at which longitude is defined to be 0°. Together, a prime meridian and its anti-meridian (the 180th meridian in a 360°-system) form a great circle. This great circle divides a spheroid into two hemispheres.

4 0
2 years ago
A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the
horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution :  Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 Cm^{3}

First, find the Density of given substance.

Formula used :    

Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}

Now,put all the values in this formula, we get

Density=\frac{46.9 g}{3.5 Cm^{3} } = 13.4 g/Cm^{3}

So, we conclude that the density of given substance (13.4 g/Cm^{3}) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/Cm^{3} respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0
3 years ago
Read 2 more answers
I need help in this :(
alexandr402 [8]

Answer:

ok follow me twitter dm me for answer The Kid D is my twitter

Explanation:

6 0
3 years ago
What information is need to calculate the percent composition of a compound?
Anestetic [448]

Answer:

Molecular formula

Explanation:

Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.

Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.

E. g., calculate the percent composition of water:

  • molecular formula is H_2O;
  • calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
  • find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
  • find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.
8 0
3 years ago
What was the velocity over the entire trip? <br><br> A. -2<br> B. -1<br> C. 0<br> D. 1<br> E. 2
zhannawk [14.2K]
You’re right its A -2
5 0
2 years ago
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