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vladimir1956 [14]
3 years ago
12

Which of the following only takes into account distance and time? Question 18 options: displacement velocity speed acceleration

Chemistry
1 answer:
MakcuM [25]3 years ago
4 0
I think its acceleration that is what i choosen on the test 
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Butane, ch3-ch2-ch2-ch3, is a hydrocarbon fuel used in lighters. how many moles of molecules are there in a 350. gram sample of
Galina-37 [17]
<span>6.03 moles.



1 molecule of butane contains 4 carbon atoms and ten hydrogen atoms.

The molar mass is 4 times the atomic mass of carbon, 12 g/mol, plus 10 times the atomic weight of hydrogen, 1 g/mol.

Molar mass = 4 * 12 g/mol + 10 * 1 g/mol = 58 g/mol.

This means that 1 mole of butane has a mass of 58 g.

To figure out how many moles are in a sample of butane, divide the mass of sample in grams by 58 grams

Number of moles in sample = 350 g / 58 g/mol = 6.03 moles.</span>
3 0
2 years ago
What is the break down of food into energy
AnnyKZ [126]

it is nutrients that's it


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3 years ago
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Ronch [10]

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8 0
3 years ago
The gas in an aerosol can is at a pressure of 3.16 atm at 32.2°C. What would the gas pressure in the can be at 22.9°C?
olasank [31]

Answer:

The pressure of the gas would be 3.06 atm

Explanation:

Amonton's law states that the pressure is directly proportional to the absolute temperature of a gas under constant volume. The equation is:

P1 / T1 = P2 / T2

<em>Where P1 is the initial pressure = 3.16atm</em>

<em>T1 is initial absolute temperature = 273.15 + 32.2°C = 305.35K</em>

<em>P2 is our incognite</em>

<em>And T2 is = 273.15 + 22.9°C = 296.05K</em>

<em />

Replacing:

3.16atm / 305.35K = P2 / 296.05K

3.06 atm = P2

<h3>The pressure of the gas would be 3.06 atm</h3>
5 0
3 years ago
What is the enthalpy for reaction 1 reversed? reaction 1 reversed: co2→co + 12o2 express your answer numerically in kilojoules p
QveST [7]
The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:

CO2: -393.5 kJ/mol
CO: -99 kJ/mol
O2: 0 kJ/mol

As observed O2 will not have enthalpy of formation as it is a pure substance. 

To calculate for the enthalpy of reaction,
   enthalpy of formation of products - enthalpy of formation of reactants
= (-99 kJ/mol) - (-393.5 kJ/mol)
    = 294.5 kJ/mol

ANSWER: 294.5 kJ/mol
8 0
3 years ago
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