In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
The average pea weighs between 0.1 and 0.36 grams.
If we take the lower value (0.1 g/pea), the number of peas in 454 g is:
If we take the higher value (0.36 g/pea), the number of peas in 454 g is:
In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
You can learn more about conversion factors here: brainly.com/question/1844638
Answer:
a) V air/day = 8640 L air an adult breaths / day
b) 0.0181 L CO intake a person / day
Explanation:
a) one average person has 12 breaths for min:
in each breath it take an average of 500 mL on air.
⇒ 12 breath / min * 500mL air / breath = 6000 mL air / min
the average air volume per day of a person is:
⇒ Vair/day = 6000 mL air / min * (60 min / h) * ( 24 h / day ) = 8640000 mLair / day * ( L / 1000 mL)
⇒ V air / day = 8640 L / day
b) 2.1 E-6 L CO / L air * 8640 L air / day = 0.0181 L CO / day
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
Answer:
A channel in an ancient Martian "river bed" was not carved out by liquid water but built by molten lava
