First let us calculate for the moles of CH3OH formed:
moles CH3OH = 23 g / (32 g / mol) = 0.71875 mol
We see that there are 2 moles of H2 per mole of CH3OH, so:
moles H2 = 0.71875 mol * 2 = 1.4375 mol
Assuming ideal gas behaviour, we use the formula:
PV = nRT
V = nRT / P
V = 1.4375 mol * (62.36367 L mmHg / mol K) * (90 + 237.15
K) / 756 mm Hg
<span>V = 43.06 Liters</span>
Answer:
a) Frope= 71.7 N
b) Frope=6.7 N
Explanation:
In the figure the skier is simulated as an object, "a box".
a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:
∑F=0
Frope=w*sen6.8°
Frope=71.71N
Take into account that w is the weight that is calculated as mass per gravitiy constant:
w=m*g
b) In this case the system has an acceleration of 0.109m/s2. Then, we apply Newton's second law of motion:
F=m*a
F=61.8Kg*0.109m/s2
Frope=6.73N
Answer:
3.036×10⁻¹⁰ N
Explanation:
From newton's law of universal gravitation,
F = Gm1m2/r² .............................. Equation 1
Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.
G = gravitational constant
Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²
Substituting into equation 1
F = 6.67×10⁻¹¹×7.9×6.1/2²
F = 321.427×10⁻¹¹/4
F = 30.36×10⁻¹¹
F = 3.036×10⁻¹⁰ N
Hence the force between the balls = 3.036×10⁻¹⁰ N
Answer:
The block will not move.
Explanation:
We'll begin by calculating the frictional force. This can be obtained as follow:
Coefficient of friction (µ) = 0.6
Mass of block (m) = 3 Kg
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (R) = mg = 3 × 10 = 30 N
Frictional force (Fբ) =?
Fբ = µR
Fբ = 0.6 × 30
Fբ = 18 N
From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.
