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Korvikt [17]
2 years ago
5

Where do you find the reactants in a chemical equation ​

Chemistry
1 answer:
Alexxandr [17]2 years ago
5 0
On the left side of the equation is the reactants (starting side) and the right side (end side) is the products.
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<h2>Answer : By weighing the costs and benefits of an environmental issue </h2><h3> Explanation :</h3>

The law makers usually conduct many studies before a law is imposed. They try to explore many other options available to the current environmental issue and then come to a conclusion to make a law.

They also weigh the cost aspect along with the benefit of the ongoing environmental issue. They try to come up with something which appears to be cost effective and result bearing.

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Why do plants need sap
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3 years ago
Read 2 more answers
Given the balanced equation:
Aliun [14]

Answer:

0.4

Explanation:

Given data:

Number of moles of SrCl₂ consumed = ?

Mass of ZnCl₂ produced = 54 g

Solution:

Chemical equation:

ZnSO₄ + SrCl₂    →    SrSO₄ +  ZnCl₂

Number of moles of ZnCl₂:

Number of moles = mass/ molar mass

Number of moles = 54 g/136.3 g/mol

Number of moles = 0.4 mol

Now we will compare the moles of  ZnCl₂ with SrCl₂  from balance chemical equation.

                          ZnCl₂              :             SrCl₂

                              1                  :                1

                           0.4                 :              0.4

Thus when 54 g of  ZnCl₂ produced 0.4 moles of SrCl₂ react.

6 0
3 years ago
The ion MnOis often used to analyze for the Fe2+ content of an aqueous solution by using the (unbalanced) reaction Mno+Fe2+ + Fe
dusya [7]

Answer:

B) 0.230 M

Explanation:

The first step is to <u>balance the reaction</u> between the Ferrous ion and the permanganate ion:

5~Fe^+^2~+~MnO_4^-^1~+~8H^+~->~5Fe^+^3~+~Mn^+^2~+4H_2O

Then we have to <u>calculate the moles</u> of MnO_4^-:

M~=~\frac{mol}{L}

mol~=~M*L

mol~=~0.033~M*0.0633L=~0.002088~mol~MnO_4^-

Then using the <u>molar ratio</u> we can find the moles of Fe^+^2:

0.002088~mol~MnO_4^-\frac{5~mol~Fe^+^2}{1~mol~MnO_4^-}=0.01044~mol~Fe^+^2

Finally we can calculate the molarity:

M=\frac{0.01044~mol~Fe^+^2}{0.0455~L}=0.230~M

5 0
3 years ago
Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con
Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is K_a\times K_b

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)

K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}

b) PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)

K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

For overall reaction on adding a and b we get c

c) P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)

K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}

K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

The equilibrium constant for the overall reaction is K_a\times K_b

4 0
3 years ago
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