Historically, the first chemical definition of an acid and a base was put forward by Svante Arrhenius, a Swedish chemist, in 1884. An Arrhenius acid is a compound that increases the H+ion concentration in aqueous solution. The H+ ion is just a bare proton, and it is rather clear that bare protons are not floating around in an aqueous solution. Instead, chemistry has defined the hydronium ion(H3O+) as the actual chemical species that represents an H+ion. H+ ions and H3O+ ions are often considered interchangeable when writing chemical equations (although a properly balanced chemical equation should also include the additional H2O). Classic Arrhenius acids can be considered ionic compounds in which H+ is the cation. Table 12.1 "Some Arrhenius Acids"lists some Arrhenius acids and their names.
contains hydrogen atoms. This is answer
The correct answer is A. <span>167 milliliters of 7% solution and 333 milliliters of 4% solution and here is how:
</span><span>If x is the number of milliliters of the 7% saline solution and y is the number of milliliters of the 4% saline solution then add up to 500 milliliters total, so x + y = 500.
</span>and if we do
x + y = 500
<span>x = 500 - y </span>
<span>0.07x + 0.04y = 25 (substitute 500 - y for x) </span>
<span>0.07(500 - y) + 0.04y = 25 </span>
<span>35 - 0.07y + 0.04y = 25 </span>
<span>-0.03y + 35 = 25 </span>
<span>-0.03y = -10 </span>
<span>y = 333.333... </span>
<span>y = about 333 </span>
<span>x = 500 - y = 500 - 333 = 167
</span>Then you know why the answer is A.
Cześć, nie mówię po polsku, więc zrobiłem to w Tłumaczu Google, więc przepraszam, jeśli coś brzmi śmiesznie.
chlor (VII) i tlen - ten wzór to 
, a tlenek chloru jest bezwodnikiem kwasu nadchlorowego.
węgiel i wodór - wzór na węgiel i wodór to CnH2n + 2), jest to związek organiczny, a niższa klasyfikacja jest taka, że jest to również węglowodór aromatyczny.
Mam nadzieję, że to pomoże, błogosławionego i cudownego dnia! :-)
-Cutiepatutie
Answer:
The solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Explanation:
Step 1: Given data
The volume of water sample = 46.0 mL.
The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.
Step 2: Calculate the solubility of X in water
46.00 mL of water sample contains 0.87 g of the mineral compound X.
To calulate how many grams of the mineral compound 1.0 mL of water sample contains:
0.87 g/46.0 mL = 0.0189 g.
This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.
