Answer:
89.53 North
Explanation:
Total N-S displacement
23.72 - 39.25 + 105.06 = + 89.53
Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.
1. Avogadro's hypothesis. Avogadro hypothesized that equal volumes of all gases (at the same pressure) will have the same number of molecules. From PV=nRT, we know that one mole of gas takes up 22.4 L
2. Mass number. The mass number is the sum of the protons and neutrons in the nucleus so Carbon 12 has an atomic number of 6 which indicates 6 protons, and a mass number of 12 so 12-6 = 6 neutrons.
3. Avogadro's number. Avogadro's number is the number of units in one mole of any substance, which has been defined as 6.02 x10^23
4. Isotopes are the different forms of a single element. They differ in neutrons. One example is Hydrogen which has three isotopes Protium, Deuterium, and Tritium.
5. Atomic mass. The mass of the atom is equal to the sum of the protons and the neutrons as electrons are so small their mass is negligible. This is not exactly the same as the mass number because this number takes into account the different isotopes
6. mole A mole has the same number of entities as 12 grams of carbon 12, it is expressed by Avogadro's number so 1 mole = 6.02 x10^23 atoms or molecules, etc
7. molar mass- the amount that one mole of substance weighs. For carbon 12, 12 grams has one mole of atoms by definition. So for carbon 12, the molar mass is 12 g/mol
It is B, Microcephaly. Hope I helped! :))
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
