Answer:
H2SO3 + 2CsOH —> Cs2SO3 + 2H2O
Explanation:
When sulfurous acid react with caesium hydroxide, caesium sulfite and water are formed according to the equation:
H2SO3 + CsOH —> Cs2SO3 + H2O
Next, we balanced the equation by putting 2 in front of CsOH and 2 in front of H2O i.e
H2SO3 + 2CsOH —> Cs2SO3 + 2H2O
Answer:
zncl2. . . . . . . . . . . . . . . .
Answer:
1.) AgNO₃
2.) 0.563 moles AgBr
Explanation:
The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).
AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)
Molarity (M) = moles / liters
100 mL = 1 L
AgNO₃
45.0 mL / 100 = 45.0 L
1.25 M = ? moles / 0.450 L
? moles = 0.563 moles
KBr
75.0 mL / 100 = 0.750 L
0.800 M = ? moles / 0.750 L
? moles = 0.600 moles
In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.
Milk is a complex colloidal system.
Answer:
Explanation:
<u>1. Word equation:</u>
- <em>mercury(II) oxide → mercury + oxygen </em>
<u>2. Balanced molecular equation:</u>
<u>3. Mole ratio</u>
Write the ratio of the coefficients of the substances that are object of the problem:

<u>4. Calculate the number of moles of O₂(g)</u>
Use the equation for ideal gases:

<u>5. Calculate the number of moles of HgO</u>

<u>6. Convert to mass</u>
- mass = # moles × molar mass
- molar mass of HgO: 216.591g/mol
- mass = 0.315mol × 216.591g/mol = 68.3g