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3 years ago

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Write the empirical formula for given elements

1 answer:

rusak2 [61]3 years ago

3 0

Answer:

$Fe(CN)_{2}\\Fe(C_{2}H_{3}O_{2})_{2}\\\\Fe(CN)_{3}\\\\Fe(C_{2}H_{3}O_{2})_{3}$

Explanation:

$Fe^{2+}(CN^{-})_{2}--->Fe(CN)_{2}\\Fe^{2+}(C_{2}H_{3}O_{2}^{-})_{2}--->Fe(C_{2}H_{3}O_{2})_{2}\\\\Fe^{3+}(CN^{-})_{3}--->Fe(CN)_{3}\\\\Fe^{3+}(C_{2}H_{3}O_{2}^{-})_{3}--->Fe(C_{2}H_{3}O_{2})_{3}$

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Read 2 more answers

A gas has a volume of 1.75L at -23C and 150.0kPa. At what temperature would the gas occupy 1.30L at 210.0kPa?

Katen [24]

Answer:

T2 = 260 K

Explanation:

<em>Given data:</em>

P1 = 150.0 k Pa

T1 = (-23+ 273.15) K = 250.15 K

V1 = 1.75 L

P2 = 210.0 kPa

V2 = 1.30 L

<em>To find:</em>

T2 = ?

<em>Formula:</em>

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}$

<em>Calculation:</em>

T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)

T2 = 260 K

8 0

3 years ago

a sample of an oxide of antimony (sb) contain 19.75 g of antimony combine with 6.5 g of oxygen . what is the simplest formula fo

frez [133]

Explanation:

The given data is as follows.

Mass of antimony = 19.75 g

Molar mass of Sb = 121.76 g/mol

Therefore, calculate number of moles of Sb as follows.

Moles of Sb = $\frac{mass}{\text{molar mass}}$

= $\frac{19.75 g}{121.76 g/mol}$

= 0.162 mol

Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.

Moles of oxygen = $\frac{mass}{\text{molar mass}}$

= $\frac{6.5 g}{16 g/mol}$

= 0.406 mol

Hence, ratio of moles of Sb and O will be as follows

Sb : O

$\frac{0.162}{0.162} : \frac{0.406}{0.162}$

1 : 2.5

We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.

Thus, we can conclude that the empirical formula of the given oxide is $Sb_{2}O_{5}$ .

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3 years ago