Answer:
the Ray's go right through it
Explanation:
the Ray's are so small they punch right through them
This problem is simply converting the concentration from molality to molarity. Molality has units of mol solute/kg solvent, while molarity has units of mol solute/L solution.
2.24 mol H2SO4/kg H2O * (0.25806 kg H2SO4/mol H2SO4) = 0.578 kg H2SO4/kg H2O
That means the solution weighs a total of 1 kg + 0.578 kg = 1.578 kg. Then, convert it to liters using the density data:
1.578 kg * (1000g / 1kg) * (1 mL/1.135 g) = 1390 mL or 1.39 L.
Hence, the molarity is
2.24/1.39 = 1.61 M
Answer: <em>Hopefully this helps! sorry if not. :))</em>
<em></em>
<em>Speed has a greater impact on mass because its increases in velocity have an exponentially greater impact on translational kinetic energy because kinetic energy is proportional to velocity squared. Doubling an object's mass would only double its kinetic energy, however doubling its momentum would quadruple its velocity.</em>
Yes it could, but you'd have to set up the process very carefully.
I see two major challenges right away:
1). Displacement of water would not be a wise method, since rock salt
is soluble (dissolves) in water. So as soon as you start lowering it into
your graduated cylinder full of water, its volume would immediately start
to decrease. If you lowered it slowly enough, you might even measure
a volume close to zero, and when you pulled the string back out of the
water, there might be nothing left on the end of it.
So you would have to choose some other fluid besides water ... one in
which rock salt doesn't dissolve. I don't know right now what that could
be. You'd have to shop around and find one.
2). Whatever fluid you did choose, it would also have to be less dense
than rock salt. If it's more dense, then the rock salt just floats in it, and
never goes all the way under. If that happens, then you have a tough
time measuring the total volume of the lump.
So the displacement method could perhaps be used, in principle, but
it would not be easy.
Answer:
- <u><em>It is positive when the bonds of the product store more energy than those of the reactants.</em></u>
Explanation:
The <em>standard enthalpy of formation</em>, <em>ΔHf</em>, is defined as the energy required to form 1 mole of a substance from its contituent elements under standard conditions of pressure and temperature.
Then, per defintion, when the elements are already at their standard states, there is not energy involved to form them from that very state; this is, the standard enthalpy of formation of the elements in their standard states is zero.
It is not zero for the compounds in its standard state, because energy should be released or absorbed to form the compounds from their consituent elements. Thus, the first choice is false.
When the bonds of the products store more energy than the those of the reactants, the difference is:
- ΔHf = ΔHf products - ΔHf reactants > 0, meaning that ΔHf is positive. Hence, the second statement is true.
Third is false because forming the compounds may require to use (absorb) or release (produce) energy, which means that ΔHf could be positive or negative.
Fourth statement is false, because the standard state of many elements is not liquid. For example, it is required to supply energy to iron to make it liquid. Thus, the enthalpy of formation of iron in liquid state is not zero.
