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jeka94
2 years ago
7

Need help with balancing equations!

Chemistry
2 answers:
kirill [66]2 years ago
8 0

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the balanced chemical equation for given reaction ~

\sf Fe_2O_3 (s) +3 \:  CO (g) =2 \:  Fe(l) + 3\: CO_2 (g)

As we can see, the Coefficient of Fe (l) is 2

shusha [124]2 years ago
4 0

Answer:

Fe2O3 + 3CO → 2Fe + 3CO2

Explanation:

the numbers in front are the numbers you need

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What is the answer to this chemistry question?
olga_2 [115]

Answer:

The answer to your question is remplacement double

Explanation:

Data

Lead (II) nitrate = Pb(NO₃)₂

Potassium iodide = KI

Process

1.- Write the balanced chemical reaction

             Pb(NO₃)₂  +  2KI  ⇒   PbI₂  +  2KNO₃

2.- Conclusion

This is a remplacement double reaction because there are two reactants that interchange cations and the products are a combination of the reactants.

5 0
3 years ago
Suppose you wish to apply SSA to a triangle, in order to find an angle
emmasim [6.3K]
First of all there nothing exists something like SSA congruence criterion, It is SAS Congruence where the sides of a Triangle and the angle between them is exactly same to the other two sides and angle between them of another triangle.

So the correct option is D.

Explanation:

If the Triangle has two sides equal, so the two angles there are equal. Also the other angle be some other definite angle. So

45+x+x=180(Considering definite angle to be 45)
2x=135
x=67.5

So there is only one solution to the triangle.
8 0
3 years ago
A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution
Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

<u>Step 1:</u> Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

<u>Step 2:</u> Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

<u>Step 3:</u> Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

<u>Step 4:</u> Calculate change in internal energy

ΔU =  Q + W       W = 0  (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0
3 years ago
Determine the enthalpy for this reaction: Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
ivanzaharov [21]
<span>Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 </span><span>kJ/mol
Al(OH)3(s)= −1277.0 </span><span><span>kJ/mol
</span> H2O(l)= −285.8 </span><span>kJ/mol
AlCl3(s) =−705.6 </span><span>kJ/mol

</span><span>Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
    reactants                         products

products- reactants:</span><span>

(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 </span>kJ per mole at 25°C
<span>
</span>






6 0
2 years ago
What family has 4 valence electrons?
PilotLPTM [1.2K]

Answer:carbon group

Explanation:

4 0
2 years ago
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