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jarptica [38.1K]
4 years ago
14

CHEMISTRY PLEASE HELP ILL GIVE U 20

Chemistry
1 answer:
dybincka [34]4 years ago
5 0
The answer is c hope it helps
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How would you pick the right kind of fluid to use in your hydraulic lift?
love history [14]

Answer:you would pick the fluid by seeing what would work such as gas in a truck or car you need to know what would work best and the best power for the lift

Explanation:

3 0
3 years ago
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Given the chemical equation: Fe2O3 + 3CO --> 2Fe + 3CO2
iren [92.7K]

Answer:

1 mole

Explanation:

m(Fe) = 112 g

n(Fe₂O₃) = ?

First you must calculate how many moles of Fe is in 112 g of Fe:

n(Fe) = m(Fe) / Ar(Fe) = 112 g / 55,85 g moles⁻¹ = 2 moles

Now, from reaction you compare moles of Fe and Fe₂O₃ :

2 n(Fe₂O₃) =  n(Fe)

n(Fe₂O₃) = n(Fe) / 2 = 1 mole

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3 years ago
Para que serve o balanceamento de equações químicas?
Marrrta [24]
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3 years ago
__Pb(NO3)2 + __NaCI = __NaNO3 + __PbCI2
Elena L [17]

Answer:

Pb(NO3)2 + 2NaCl = 2NaNO3 + PbCl2

6 0
2 years ago
The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
LiRa [457]

Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.

Hence, equation for rate law of reaction will be as follows.

              Rate = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
3 years ago
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