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stepan [7]
4 years ago
6

I’m driving down the highway at 75.0 miles/hour, how fast is that in m/seconds?

Chemistry
1 answer:
Aliun [14]4 years ago
4 0

Answer:

1.25 miles/ 1 minute

60x1.25=7.50

Explanation:

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I NEED HELP QUICK! BRAINLIEST TO WHO IS RIGHT!
Travka [436]

Answer:

3rd option

Explanation:

3 0
3 years ago
How do I get an atomic mass
ahrayia [7]
To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons.
3 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
Given that you added about 5ug of purified acid phosphatase to tube A, calculate the amount of acid phosphatase that was present
Reil [10]

Hey there!:

Amount of purified acid phosphatase added to tube A = 5 ug ( micrograms )

Amount of acid phosphatase present in 400 ug of wheat germ extract  in tube B :

1 / 100 = 5x / 400 =

100 x = ( 0.5 ) ( 400 ) =

x = (0.5 ) ( 400 ) = 100

x = 200 / 100

x =  2 ug


The amount of acid phosphatase present in 400 ug of the  wheat germ in tube B is 2 ug


Hope that helps!

6 0
4 years ago
Ration for molar volume. 1 mole=___L at STP
solmaris [256]

sorry idk!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!11


4 0
3 years ago
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