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KonstantinChe [14]
3 years ago
15

Joe the scientist wants to figure out the angle for throwing a football that

Chemistry
2 answers:
Solnce55 [7]3 years ago
7 0

Answer:

C

Explanation:

dedylja [7]3 years ago
7 0

Answer:

c

Explanation:

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Is gin a heterogeneous or homogeneous mixture?
Gelneren [198K]
Gin is uniform throughout and is a homogenous mixture. If it wasn't you would have awful lumps in your drink :). Hope I helped!
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How are the graphs of the sine function and the cosine function different?
Zielflug [23.3K]
They are different by a phase shift of pi/2
3 0
2 years ago
Select The on that most applys<br><br><br> Will mark brainliest
sdas [7]

Answer:

A and C

Explanation:

7 0
3 years ago
Calculate the theoretical carbonaceous and nitrogenous oxygen demand for:
serg [7]

Answer:

The correct answer is 129 mg and 232 mg.

Explanation:

Theoretical carbonaceous oxygen demand:

The reaction will be,  

C₂H₆O₂ + 5/2 O₂ ⇒ 2CO₂ + 3H₂O

Thus, for one mole of C₂H₆O₂ (ethylene glycol), 2.5 moles of O₂ is needed.  

The molecular mass of ethylene glycol is 62 grams per mole.  

The given mass of ethylene glycol is 100 mg or 0.1 grams

The moles of ethylene glycol will be,  

Moles = Weight/Molecular mass

= 0.1/62 = 1.613 × 10⁻³ mol

For 1.613 × 10⁻³ mol, the moles of O₂ will be,  

= 2.5×1.613×10⁻³

= 4.0.×10⁻³ × 32mol

= 0.129 grams or 129 mg.  

The theoretical nitrogenous oxygen demand is:  

The reaction will be,  

2NH₃-N + 9/2O₂ ⇒  4HNO2 + H₂O

Thus, for 2 moles of NH₃-N, 4.5 moles of O₂ is needed,  

Therefore, for 1 mol of NH₃-N, the oxygen required will be,  

= 4.5/2 = 2.25 mol

The given mass of NH₃-N is 100 mg, the moles of NH₃-N will be,  

Moles = 100×10⁻³/31 = 3.225 × 10⁻³ mol (The molecular mass of NH₃-N is 31 gram per mole)

The moles of O₂ is 2.25 × 3.225 × 10⁻³ = 7.258 × 10⁻³ mol.  

Now the mass of O2 will be,  

= 7.258 × 10⁻³ × 32

= 0.232 grams

= 232 mg

5 0
3 years ago
Calculate the second volumes 980L at 71C and 107.2atm to 13C and 59.3atm
Lapatulllka [165]

Answer:

V₂ = 1473.03 L

Explanation:

Given data:

Initial volume = 980 L

Initial pressure = 107.2 atm

Initial temperature = 71 °C (71 +273.15 = 344.15 K)

Final temperature = 13°C (13+273.15 = 286.15K)

Final volume = ?

Final pressure = 59.3 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 107.2 atm × 980L × 286.15 K / 344.15 K×   59.3 atm

V₂ = 30061774.4 atm .L. K / 20408.095 atm. K

V₂ = 1473.03 L

5 0
3 years ago
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