0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol
mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol
1 mol --------- 164g
0,06125 ---- X
X = 10,045g
To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
<span>The part of making a solution that always releases energy is the overall change in forming the solution. The answer is letter D. Although letters A, B and C can be viable answers but, it is not always the case. There are some substances that when you mix or separate them requires more energy or less energy. An example would be w</span>hen the formation (or enthalpy of formation) of carbon
dioxide is negative, it means that it releases heat to the surroundings. When
it releases heat to the surroundings, the reaction is exothermic. Another example is when you mix baking soda and muriatic acid, the resulting mixture is colder. When it is cold, it means that the reaction is endothermic. So the best answer is letter D.
1.) A compound is a thing that is composed of two or more (separate) elements.
2.) A compound cannot be separated easily, while its properties differ from where it originates from. It also is formed from chemical reactions.
3.) A pure substance, also known as chemical substance, is a material with constant composition and consistant properties.
4.) If a substance is not a pure substance, then it is an impure substance.
Hope this helped. :)
Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
PH + pOH = 14
11.8 + pOH = 14
pOH = 14 - 11.8
pOH = 2.2
[OH-] = 10 ^- pOH
[OH-] = 10 ^- 2.2
[OH-] = <span>6.33 x 10^-3 M
</span>
Answer B
hope this helps!