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3 years ago

Calculate the molar mass FeCl3 Fe2(SO4)3 PbO PbO2 Pb3O4

1 answer:

8 0

To find molar mass: look for each element on the table and find their atomic masses. Then add each atomic mass to find the molar mass. If there is a subscript then that means that there is 3 of that number. For atomic mass round to the nearest hundredths
FeCl3= 162.2
Fe2(SO4)3= 354.88
PbO=223.2
PbO2 = 239.2
Pb3O4=685.6

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Would you expect manganese(II) oxide, MnO, to react more readily with HF(aq) or NaOH(aq)?

ella [17]

Answer:

NaOH(aq)

Explanation:

NaOH(aq) is known to precipitate Mn^2+ ions according to the following reaction; Mn^2+(aq)+2OH^−(aq)↽−−⇀Mn(OH)2(s)

Hence, manganese(II) oxide reacts more readily with NaOH(aq) under ordinary conditions precipitating the metal hydroxide solid. This is one of the characteristic reactions of Mn^2+.

3 0

2 years ago

Calculate the grams of so2 gas present at stp in a 5.9 l container.

nata0808 [166]

Answer: The mass of sulfur dioxide gas at STP for given amount is 16.8 g

Explanation:

At STP conditions:

22.4 L of volume is occupied by 1 mole of a gas.

So, 5.9 L of volume will be occupied by = $\frac{1mol}{22.4L}\times 5.9L=0.263mol$

Now, to calculate the mass of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sulfur dioxide gas = 0.263 mol

Molar mass of sulfur dioxide gas = 64 g/mol

Putting values in above equation, we get:

$0.263mol=\frac{\text{Mass of sulfur dioxide gas}}{64g/mol}\\\\\text{Mass of sulfur dioxide gas}=(0.263mol\times 64g/mol)=16.8g$

Hence, the mass of sulfur dioxide gas at STP for given amount is 16.8 g

7 0

3 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$