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mrs_skeptik [129]
3 years ago
8

When calculating percent error, the accepted value is in the denominator True or false

Chemistry
2 answers:
kaheart [24]3 years ago
8 0

Answer:

True

Explanation:

AVprozaik [17]3 years ago
6 0
The answer is TRUE

Have a nice day / night! :)
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A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc
Paul [167]

Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

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\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

T_1 = initial temperature of O_2 gas = 25^oC=273+25=298K

T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

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5 0
3 years ago
In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.
kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}

n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{Fe} = 0.5\:mol}

* to n(S)

n_{S} = \dfrac{m_{S}}{MM_{S}}

n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{S} = 0.5\:mol}

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

Fe + S \Longrightarrow FeS

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}

m_{FeS} = n_{FeS}*MM_{FeS}

m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}

\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}

Answer:

44 grams of iron sulfide

___________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

4 0
3 years ago
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