Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm
The answer is (1) CaCO3. The bond between C and O is covalent bond. And the bond between Ca2+ and CO3 2- ion is the ionic bond. So the compound CaCO3 satisfy this condition.
Answer:
The temperature is 12.35 C.
Explanation:
An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
- P= 2.15 atm
- V= 35.28 L
- n= 3.24 moles
- R= 0.082

- T= ?
Replacing:
2.15 atm* 35.28 L= 3.24 moles* 0.082
* T
Solving:

T=285.5 K= 12.35 C (being 273.15 K= 0 C)
<u><em>The temperature is 12.35 C.</em></u>
Answer:
Answer: 2.5 mol O2 are needed to make 5 mol H2O, assuming H2 is in excess.
Explanation:
Answer: 1.19 M
Explanation:
molarity = (moles of solute)/(liters of solution) = 9.870 / 8.26 = 1.19 M