Answer:- Oxidation number of Cl does not change as it is -1 on both sides.
Explanations:- oxidation number of Mg on reactant side is 0 as it is in its elemental form(not combined with another element).
Oxidation number of hydrogen in its compounds is +1, so if H is +1 in HCl the oxidation number of Cl is -1 as the sum has to be zero.
On product side, Mg oxidation number is +2 as the oxidation number of alkaline earth metals in their compounds is +2.
Two Cl are present in magnesium chloride, so if Mg is +2 then Cl is -1.
Oxidation number of H on product side is 0 as it is present in its elemental for,
,
So, it is only chlorine(Cl) whose oxidation number does not change for the given equation.
Seeing signs of a chemical reaction does not always mean that a reaction is happening. For example, a gas (water vapor) is given off when water boils. ... You can tell that it is a physical change because water vapor can condense to form liquid water. In a chemical change, a new substance must be produced.
Answer:
2.7 g/mL:)
An aluminum bar was found to have a mass of 27g. Using water displacement, the volume was measured to be 10 ml. What is the density of the aluminum? Group of answer choices (27 g)/(10 ml) (10 ml )/(2.70 g) (270 g)/(10 ml) (10 ml )/(27 g)
Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
Answer:
Please, see attached two figures:
- The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.
- The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.
Explanation:
The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the <em>solubility</em> of <em>NH₄Cl.</em>
From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.
Assuming density 1.0 g/mol for water, 10 mL of water is:
Thus, the solutibily is:
