For many solids<span> dissolved in </span>liquid<span> water, the </span>solubility <span>increases with </span>temperature<span>.</span>
Answer:
2.835 moles of carbon
Explanation:
By definition, there are 6.022x10^23 atoms (or compounds) in one mole.
Write and use this as a conversion factor:
(6.022x10^23 atoms)/mole
(1.707 x 10^24 atoms of carbon)/((6.022x10^23 atoms)/mole) = 2.835 moles carbon
This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.
<h3>Molarity</h3>
In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

That in milliliters turns out to be:

Learn more about molarity: brainly.com/question/10053901
An electrode is an electrical conductor used to make contact with a nonmetallic part of a circuit The word was coined by William Wheel at the request of the scientist Michael Faraday from the Greek words electron, meaning amber and hods, a way.
Answer:
The base must be ethylamine.
Explanation:
The pH of solution of a weak base gives us an idea about the Kb of the base.


![pOH=-log[OH^{-}]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E%7B-%7D%5D)
![[OH^{-}]=0.0062M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D0.0062M)
The relation between Kb and hydroxide ion concentration is:
![Kb=\frac{[OH^{-}]^{2}}{[base]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BOH%5E%7B-%7D%5D%5E%7B2%7D%7D%7B%5Bbase%5D%7D)

Thus the weak base must be ethylamine.