Ask question

Ask question

All categories

3 years ago

13

Determine the name of the following compound Fe(OH) 3 , Provide explanation of how you determined the name, including the type o

f compound this , Is this compound an acld or a base?

1 answer:

stepan [7]3 years ago

7 0

Answer:

1. Iron(iii) hydroxide

2. It is a base

Explanation:

To determine the name of the compound, we must find the oxidation number Fe in the compound since the oxidation number of O is —2 in all its compound except for peroxide where it —1 and the oxidation number of Hydrogen is always +1 in all its compounds except for hydrides where it is —1. The oxidation number of Fe can calculated as follows:

Fe(OH)3 = 0

O = —2

H = +1

Fe =?

Fe + 3( —2 + 1) = 0

Fe + 3(—1) = 0

Fe —3 = 0

Fe = 3

The oxidation number of Fe in Fe(OH)3 is 3.

Therefore, the name of Fe(OH)3 is Iron(iii) hydroxide

Fe(OH)3 is a base since it contains the OH group

You might be interested in

4.00 g of O2 gas are in a sealed, 2.00 L gas canister at 22.0 °C what is the pressure inside this container (in atm)?

Volgvan

Answer:

1.51448 atms

Explanation:

6 0

3 years ago

A cold front in New Hampshire in December will bring __________.

JulijaS [17]

i really dont know and im so so so so sorry

6 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago

Can someone help Plz

Marrrta [24]

Answer: A. 1.60 liters

Explanation:

$2Na_2O_2+2CO_2\rightarrow 2Na_2CO_3+O_2$

As can be seen from the balanced chemical equation, 2 moles of $CO_2$ produce 1 mole of $O_2$ .

According to Avogadro's law, every 1 mole of the gas occupies 22.4 liters at STP.

Thus $2$ moles of $CO_2$ occupies $22.4\times 2=44.8L$ at STP and produce 1 mole of $O_2$ i.e. 22.4 L

44.8 L of $CO_2$ will produce 22.4 L of $O_2$

Thus $3.20L$ of $CO_2$ will produce $\frac{22.4}{44.8}\times 3.20=1.60L$

6 0

3 years ago

A single atom of an element has 21 nutrons 20 electtons and 20 protons which element is it

guajiro [1.7K]

your element would be calcium

7 0

3 years ago