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What mass, in grams, of fluorine has (F2) is contained in a 10.3-liter tank at 29.8 degrees Celsius and 1.23 atmospheres? Show a

denis23 [38]

PV = nRT
P is pressure, V is volume in L, n is number of moles, R is the gas constant, and T is temperature in K

(1.23 atm)(10.3 L) = (n)(.08206)(29.8 + 273)

n = .5 moles x 38 grams per mol F2 = 19.4 grams F2

8 0

2 years ago

A firecracker exploding would be an example of a (n) ___? A. Physical change B. Exothermic reaction C. Formation of a precipitat

Jobisdone [24]

The answer would be A) or the first option.

3 0

3 years ago

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Lab: Limiting Reactant and Percent Yield

creativ13 [48]

Answer : If a substance is the limiting reactant, then it limits the formation of products because in the reaction it is present in limited amount.

Explanation :

While observing a chemical reaction, we can tell about whether a reactant is limiting or excess.

Step 1 : first write the chemical reaction and then balanced the chemical equation.

$C_2H_4+H_2\rightarrow C_2H_6$

Step 2 : convert the given masses into the moles if mass of $C_2H_4$ is 10.5 g and molar mass of $C_2H_4$ is 28 g/mole and the mass of hydrogen is 0.40 g and molar mass of hydrogen is 2 g/mole.

$\text{moles of }C_{2}H_{4}=\frac{10.5g}{28g/mole}=0.375moles$

$\text{moles of }H_{2}=\frac{0.40g}{2g/mole}=0.20moles$

Step 3 : Now we have to determine the limiting reagent and excess reagent.

$\text{ moles of }C_{2}H_{4}\text{ in excess}=0.375-0.20=0.175\text{moles}$

Now we conclude that $C_{2}H_{4}$ is the limiting reagent and hydrogen is an excess reagent.

Hypothesis :

Limiting reagent : It is the reagent in the chemical reaction that is totally consumed when the chemical reaction is complete. Limiting reagent limits the formation of products.

7 0

2 years ago

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When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for

balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

$Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-$

Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$