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: Your code returns a number of 99.123456789 +0.00455679
Ok, you must see where the error starts to affect your number.
In this case, is in the third decimal.
So you will write 99.123 +- 0.004 da da da.
But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.
in the error, after the 4 comes a 5, so it rounds up.
So the final presentation will be 99.123 +- 0.005
you are discarding all the other decimals because the error "domains" them.
Answer:
B) with 9/10 submerged
Explanation:
= mass of ice cube
= density of soft drink
= Volume of soft drink displaced
ice cube floats in the soft drink when the force of buoyancy on it balances its weight. Force of buoyancy acting on the cube in upward direction is same as the weight of the soft drink displaced. hence we can write
weight of ice cube = weight of soft drink displaced
we see that the acceleration due to gravity cancel out both side and hence it does affect as astronaut is on earth on in a lunar module.
Explanation:
Basaltic lava
Basaltic lava generally takes two distinct forms known by the Hawaiian terms pahoehoe and aa. Pahoehoe has a smooth wavy surface that resembles twisted rope. It advances by extruding molten toes of lava beneath a thin, flexible crust. As it travels pahoehoe lava often changes to blocky flows called aa.
Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K
