Answer:
59.76 grams of Li₂O
Explanation:
To solve this problem we will first calculate the moles of Li₂O. One mole of given substance always contains 6.022 × 10²³ particles which can be atoms, ions, molecules or formula units. This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Particles ÷ 6.022 × 10²³
Putting values,
Number of Moles = 1.204 × 10²⁴ Particles ÷ 6.022 × 10²³
Number of Moles = 2.0 Moles
Secondly, we will convert calculated moles to mass using following relationship.
Moles = Mass / M.Mass
Or,
Mass = Moles × M.Mass
Putting values,
Mass = 2.0 mol × 29.88 g/mol
Mass = 59.76 grams of Li₂O
The solution for this problem is:
Let x = speed of wind
Speed of plane with the wind = x + 100
Speed of plane against the wind = 100 -x
We will be using the formula for distance which is (Rate)(Time), getting the formula for time would be distance/rate Time to travel 600 miles with the wind = Time to travel 400 miles against the wind 600/(x + 100) = 400/(100 - x)
400(x + 100) = 600(100 - x)
400x + 40000 = 60000 - 600x
1000x = 20000
x = 20000/1000
x = 20 mph
Answer:
<u>167.2 g</u>
Explanation:
Known
VC4H10 = 21.3
T = 0.00 C (convert to Kelvin: 273 K)
P = 1.00 atm
Unknown
m = ?g
1. <u>Write the balanced chemical equation</u>
1 C4H10 + 1O2 -----> 4 CO2 + 5 H2O
2. <u>Find the volume ratio of Carbon Dioxide to Butane </u>
1 C4H10 4 CO2 = 4 volumes CO2 / 1 volume C4H10
3.<u> Multiply by the known volume of n (butane)</u>
21.3 L C4H10 x 4 volumes CO2 / 1 volume C4H10 = 85.2 L C4H10
4. <u>Use ideal gas law</u>
PV = nRT solve for n ----> n = PV/RT
n= (1.00 atm) (85.2 L) / (0.0821 L atm/mol K) (273) = 3.80 mol CO2
5.<u> Find molar mass of CO2</u>
1 C x 12 + 2 O x 16 = 44.00
6. <u>Multiply the ideal gas law solution (3.80) by molar mass CO2 (44.00)</u>
3.80 mol CO2 x 44.00 g CO2
= 167.2 g CO2
Answer:
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