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Rainbow [258]
3 years ago
5

A small cube of iron and a large flat sheet of iron contain the same volume. Which one will completely rust first? Explain why.

Chemistry
2 answers:
Y_Kistochka [10]3 years ago
8 0

Answer:

The flat sheet will completely rust before the iron cube. Since they both have the same volume, the flat sheet has more surface area than the small cube. This means more particles are exposed on the flat sheet that can react in a chemical reaction.

Hunter-Best [27]3 years ago
4 0

The flat sheet will completely rust before the iron cube. Since they both have the same volume, the flat sheet has more surface area than the small cube. This means more particles are exposed on the flat sheet that can react in a chemical reaction.

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Percent yield is the quantity of product actually produced compared to the quantity _______.
ss7ja [257]

Answer:

Of expected product

Explanation:

6 0
2 years ago
What is the value of x in the equation below?
Colt1911 [192]
If the question includes x only than it’s value would be 1
8 0
3 years ago
Read 2 more answers
What is the limiting reactant in a chemical reaction?
IgorC [24]

Answer:

D.) the reactants that runs out first

Explanation:

Limiting reactant is the reactant which is present in the smallest amount and thus limit the yield of product.

In any chemical reaction limiting reactant is identified by steps:

First we will calculate the number of moles of given amount of reactants.

Then we will find the number of moles of product by comparing with moles of reactant through balanced chemical equation.

Then we will identified the reactant which produced smaller amount of product.

It can be better understand by following problem.

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Which is limiting reactant= ?

Chemical equation:

CaCO₃ + 2HCl  → CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                   CaCO₃         :               CaCl₂

                       1               :               1

                     0.25           :            0.25

                   HCl              :                CaCl₂

                     2                :                 1

                   0.36            :               1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of calcium chloride:

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

5 0
2 years ago
Pls help meeee!!!!!!!!!!!!!!!
Verdich [7]

Answer:

i think it's either A or D.

3 0
3 years ago
Ethanol melts at -114 degree C. The enthalpy of fusion
Brut [27]

Answer: The heat required is 6.88 kJ.

Explanation:

The conversions involved in this process are :

(1):ethanol(s)(-135^0C)\rightarrow ethanol(s)(-114^0C)\\\\(2):ethanol(s)(-114^0C)\rightarrow ethanol(l)(-114^0C)\\\\(3):ethanol(l)(-114^0C)\rightarrow ethanol(l)(-50^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of ethanol = 25.0 g

c_{p,s} = specific heat of solid ethanol= 0.97 J/gK

c_{p,l} = specific heat of liquid ethanol = 2.31 J/gK

n = number of moles of ethanol = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{25.0g}{46g/mole}=0.543mole

\Delta H_{fusion} = enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole

T_{final}-T_{initial}=\Delta T = change in temperature

The value of change in temperature always same in Kelvin and degree Celsius.

Now put all the given values in the above expression, we get

\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]

\Delta H=6885.93J=6.88kJ     (1 KJ = 1000 J)

Therefore, the heat required is 6.88 kJ

3 0
2 years ago
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