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4 years ago

Write the empirical formula for at least four ionic compounds that could be formed from the following ions:

Chemistry

1 answer:

love history [14]4 years ago

4 0

Answer:

$FeSO_{4},FeCH_{3}CO,Fe_{2}(SO_{4})_{3},Fe_{2}(CH_{3}CO)_{3}$

Explanation:

Empirical formula of ionic compound formed by two ions $A^{x+}$ and $B^{y-}$ is $A_{y}B_{x}$ (for $x\neq y$ ) of AB (for x = y)

The above empirical formula is in accordance with charge neutrality principle

Here each cation ( $Fe^{3+}$ and $Fe^{2+}$ ) can form two ionic compounds by combining with two given anions ( $SO_{4}^{2-}$ and $CH_{3}CO^{2-}$ ).

So the four ionic compounds are: $FeSO_{4},FeCH_{3}CO,Fe_{2}(SO_{4})_{3},Fe_{2}(CH_{3}CO)_{3}$

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A displacement reaction takes place when calcium is placed in a solution of zinc chloride. Which substances will most likely be

Karolina [17]

Hi!

The correct option would be A.

This is because the displacement reaction would take place as follows

Ca + ZnCl2 --> CaCl2 + Zn

A displacement reaction is one in which a substitution occurs, as the more reactive element in the mixture replaces one that is less reactive.

In the electrochemical series, we find Ca higher than Zn, which is indicative of Ca being more reactive, and having the capacity to displace Zn to form a compound.

Option D would be incorrect as no such substitution occurs.

Option B would be incorrect because again, there is no substitution occurring, and also because two metals alone (Ca and Zn in our case) can never react to form a compound.

Option C would be incorrect because it is not possible because CaCl and ZnCl are forms that are too unstable to exist due to an overall positive charge.

Hope this helps!

5 0

3 years ago

Se sabe que 10 g de calcio reaccionan con 4 g de oxígeno para obtener 14 g de óxido de calcio. Indica la cantidad de óxido de ca

egoroff_w [7]

Answer:

Si se usan 50 gramos de calcio y óxigeno, se obtienen 70 gramos de óxido de calcio.

Explanation:

Hola,

En este caso, la reacción llevada a cabo es:

$2Ca+O_2\rightarrow 2CaO$

De este modo si asumimos el ejemplo dado, 50 gramos de calcio, cuya masa atómica es 40 g/mol y 50 g de oxígeno, cuya masa atómica como gas diatómico es 32 g/mol, antes de calcular los gramos de óxido de calcio producidos, debemos identificar el reactivo límite. Así, calculamos las moles de calcio disponibles en 50 g:

$mol_{Ca}^{disponible}=50gCa*\frac{1molCa}{40gCa} =1.25molCa$

Y también las moles de calcio consumidas por los 50 g de oxígeno, utilizando su relación molar 2:1:

$mol_{Ca}^{consumidas\ por\ O_2}=50gO_2*\frac{1molO_2}{32gO_2} *\frac{2molCa}{1molO_2} =3.125molCa$

Por lo tanto, hay menos calcio disponible que el que consume el oxígeno, por lo que el calcio esel reactivo límite. Ahora, con este, calculamos los gramos de óxido de calcio, cuya masa molar es 56 g/mol, que se producen:

$m_{CaO}=1.25molCa*\frac{2molCaO}{2molCa}* \frac{56gCaO}{1molCaO}\\ \\m_{CaO}=70gCaO$

Esto quiere decir que de 50 gramos de oxígeno, solo 20 gramos reaccionan para formar 70 gramos de óxido de calcio.

Saludos!

4 0

3 years ago

Read 2 more answers

Name the reactants and products of Na3PO4 + 3KOHàNaOH + K3PO4 (Question 2 and 3)

Leona [35]

Answer:

Question 2: Na3PO4, KOH; Question 3: Na3PO4, KOH

Explanation:

Question 2

The reactants in a chemical equation are the species on the left side of the reaction arrow.

Thus the reactants are Na3PO4, KOH (sodium phosphate and potassium hydroxide).

Question 3.

The products in a chemical equation are the species on the right side of the reaction arrow.

Thus the products are NaOH, K3PO4 (sodium hydroxide and potassium phosphate).

6 0

3 years ago

Question 8 of 8> 0 Attempt 2 A chemist needs to determine the concentration of a sulfuric acid solution by titration with a s

Nesterboy [21]

Answer:

12.29 M

Explanation:

The reaction that takes place is:

H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O

Now let's calculate the moles of H₂SO₄ that were titrated:

$0.1284 M * 0.1915L*\frac{1molAcid}{2molNaOH}$ = 0.01229 mol H₂SO₄.

Thus, the concentration of the diluted solution is:

0.01229 mol H₂SO₄ / 0.010 L = 1.229 M

Finally, the concentration of the original acid solution is:

$1.229 M*\frac{250mL}{25mL}$ = 12.29 M

5 0

3 years ago

General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.

Helen [10]

Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

Q: heat absorbed

C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)

ΔT: temperature change (2.29 °C)

Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ

According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

We will use the conversion factor 1 Cal = 4.186 kJ.

86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

Step 3: Calculate the number of Cal per gram of candy

20.6 Cal/6.70 g = 3.07 Cal/g

3 0

3 years ago