Option B is correct
K = Kp /Kr
The given equation indicating, the product containing 6 moles of proton whereas the reactant contains 2 mole of bismuth and 3 mole of hydrogen sulphide.
Hence, in reaction B there are 2 mole of bismuth and 3 mole of hydrogen sulphide reacting to produce 6 moles of proton. whereas the concentration of Bi2S3 is not considered as it is present in solid phase.
Answer:
The number of mol is: 0, 042 mol in 4 grams of MgCl2
Explanation:
We calculate the weight of 1 mol of MgCl2:
Weight 1mol of MgCl2= weight Mg + (weight Cl)x 2=
24, 3 grams + 2 x 35, 5 grams = 95, 3 grams/mol MgCl2
95, 3 grams------1 mol MgCl2
4 grams -------x = (4 grams x1 mol MgCl2)/ 95, 3 grams= 0, 04197 mol MgCl2
In order to balance an equation, we apply the principle of conservation of mass, which states that mass can neither be created nor destroyed. Therefore, the mass of an element before and after a reaction remains constant. Here, the balanced equation becomes:
4Al + 3O₂ → 2Al₂O₃
The coefficients are 4, 3 and 2.
Answer;
C. unchanged rock and mineral fragments
Explanation;
A large number of landforms and features found in desert environments are formed as the result of weathering. Weathering is defined as the breakdown and deposition of rocks by weather acting in situ
The two main types of weathering which occur in deserts are Mechanical weathering, which is the disintegration of a rock by mechanical forces that do not change the rock's chemical composition and Chemical weathering, which is the decomposition of a rock by the alteration of its chemical composition.
By contrast much of the weathered debris in deserts has resulted from mechanical weathering. Chemical weathering, however, is not completely absent in deserts. Over long time spans,clays and thin soils do form.
Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.
Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
Molality = 0.1724 mol / kg
