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Morgarella [4.7K]
2 years ago
8

Write the chemical equation for ionic reaction between nabr and agno3. b boldi italicsu underline bulleted list numbered list su

perscript subscript0 words
Chemistry
1 answer:
evablogger [386]2 years ago
7 0
Step  1;  write  the  molecular  equation, that  is
NaBr(aq)  +AgNO3 (aq)---> AgBr(s) +NaNO3   the equation  is  balanced
       ionic  equation is  therefore
Na+(Aq)  +Br- (aq)  + Ag+(aq) + No3-(aq)  --->AgBr(s)  +Na+(aq)  + No3-(aq)

net  equation
Ag+(aq)  +Br-(aq) --->AgBr(s)
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Represent the formation of cations for the following metal atoms using electron dot structures. (a) Al (b) Sr (c) Ba​
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the formation of cations by using electron dot structures are :

a) Al

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Al .    losing the three valence electrons makes the Al³⁺

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b) Sr :  

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A Lewis electron dot diagram is a representation of the valence electrons of an atom that employments specks around the image of the element. The number of dots equals the number of valence electrons within the molecule. These dots are arranged to the right and left and over and underneath the symbol, with no more than two dots on a side. Cations are the positive ions shaped by the loss of one or more electrons. The foremost commonly shaped cations of the representative elements are those that include the loss of all of the valence electrons.

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3 0
1 year ago
One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
Ugo [173]

Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

8 0
3 years ago
Will give brainliest
slamgirl [31]
I will help you with answering this question.
3 0
3 years ago
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