1 significant figure, because there is no decimal after the zero the zero doesn't count.
Answer:
The temperature of a substance when the average kinetic energy of its particles increases and decreases when the average kinetic energy decreases.
Explanation:
Atoms and molecules are in constant motion. Kinetic energy is a form of energy, known as energy of motion. Kinetic energy is a form of energy, known as energy of motion. The kinetic energy of an object is that which is produced due to its movements, which depends on its mass (m) and speed (v).
Temperature refers to a quantity used to measure the kinetic energy of a system. That is, temperature is defined as an indicator of the average kinetic energy of the particles in a body.
So, since temperature is a measure of the speed with which they move, the higher the temperature the faster they move.
Finally, <u><em>the temperature of a substance when the average kinetic energy of its particles increases and decreases when the average kinetic energy decreases.</em></u>
Answer: a
Explanation:
Industry uses only about 18% while the others use around 70-90% of water.
YW!!! please mark branlest!!!! =^.^=
Answer:
mole
Explanation:
The mole in chemistry is used to represent the amount of any substance. Just like quantifying everyday things like a dozen, score, gross etc, it is a convenient unit of quantity of particles. A mole denotes 6.02 x 10²³particles of a susbstance.
Therefore, a mole is the standard unit(SI) for the amount of isopropyl alcohol in a beaker.
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
