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aleksandrvk [35]
3 years ago
6

Which correctly lists the three gases that are part of Earth's cycles in both the atmosphere and biosphere?

Chemistry
2 answers:
tiny-mole [99]3 years ago
8 0

Answer:

The correct answer is actually, B) carbon dioxide, nitrogen, oxygen

zubka84 [21]3 years ago
4 0

Answer:

B

Explanation:

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1s2 2s2 2p2 i hope this helps
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2 years ago
HELP. NO FAKE ANSWERS. I WILL REPORT. I AM CONFUSED AND NEED HELP. FILL IN THE NOT FILLED BOXES POR FAVOR.
SCORPION-xisa [38]
Question #1
Potasium hydroxide (known)
 volume used is 25 ml 
Molarity (concentration) = 0.150 M
Moles of KOH used 
           0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4) 
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l) 
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
      0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid)  concentration
    0.001875 moles × 1000/15  
        = 0.125 M

Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
      0.25 × 30/1000
        =  0.0075 moles 
Sodium Hydroxide (base)
Volume used 20 ml 
Concentration (unknown)
The equation for the reaction is 
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr   is 1 : 1
Therefore, moles of NaOH used;
                 = 0.0075 moles
NaOH concentration will be 
       = 0.0075 moles × 1000/20
       = 0.375 M

7 0
3 years ago
Describe how radioactive isotopes are used to treaat cancer​
stich3 [128]

Answer:

Radioisotope therapy is a procedure in which a liquid form of radiation is administered internally through infusion or injection. RIT's ultimate purpose is to treat cancerous cells with minimal damage to the normal surrounding tissue. These therapies are not normally the first approach used to fight a patient's cancer.

Explanation:

5 0
2 years ago
Im in class rn real school help
Ulleksa [173]

Answer:

Weathering involves the use of water and or wind to wear down the environment. The sand dunes were most likely carried by wind to form them and form its shape. Erosion most likely took away parts of the sand to make the sand dune. Deposition most likely deposited sand to the area.

Explanation:

If you possibly want a better explanation, try looking it up :).

#teamtrees #WAP (Water And Plant)

8 0
3 years ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
2 years ago
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