How do you look for elements with similar properties on the periodic table?

I WIll upvote for answer<br> Balance the following reaction Fe^2 O ^3 +HCI+FeCI^3+H^2O

kumpel [21]

Let's balance step by step. We'll start with Iron (Fe)
Fe2O3 + HCl --> FeCl3 + H2O
We have 2 Fe on the left, and only one on the right, so we'll double the Fe on the right
Fe2O3 + HCl --> 2FeCl3 + H2O
Now we have six Cl on the right, and only one on the left, so we'll multiply the Cl by six on the left
Fe2O3 + 6HCl --> 2FeCl3 + H2O
Finally, we can balance the water, as we have 6 H and 3 O on the left, and 2 H and one O on the right, so we can triple the H2O on the right
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
The equation has been balanced.

4 0

3 years ago

A compound is 52.0% zinc 9.6% carbon and 38.4% oxygen. Calculate the empirical formula of the compound.

skad [1K]

Answer:

So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.

Explanation:

What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.

6 0

1 year ago

A group of people take a diet pill. After 3 months, they measure the amount of weight they lost.

iris [78.8K]

dependent variable I think

8 0

3 years ago

Read 2 more answers

A force of 3000N is applied on a 20kg ball. Find the acceleration of the ball

Andrews [41]

Answer:

force = 3000N

mass= 20 kg

now

F= ma

3000= 20×a

3000÷20=a

a=15

F= ma by newtons second law of motion

3 0

2 years ago

Write the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by each:

Maru [420]

Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and $1s^22s^22p^6$

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1$

This element will easily loose 1 electron and form $Rb^+$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

$1s^22s^22p^3$

This element will easily gain 3 electrons and form $N^{3-}$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

$1s^22s^22p^6$

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^5$

This element will easily gain 1 electron and form $Br^-$ ion which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

4 0

2 years ago