Help me !!!!!!!!!!! Please

What is an ion? What are the two ways in which an ion can be made?

denpristay [2]

Answer: Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the ...

3 0

3 years ago

Write formula of compound given two chemicals

Evgesh-ka [11]

Nitrous oxide has 2 nitrogen atoms and 1 oxygen atom per molecule

3 0

3 years ago

What is the ion #for lodine (l)?

AveGali [126]

Answer:An iodide ion is the ion I−. Compounds with iodine in formal oxidation state −1 are called iodides. This page is for the iodide ion and its salts, not organo iodine compounds. In everyday life, iodide is most commonly encountered as a component of iodized salt, which many governments mandate.

brainliest pls

5 0

3 years ago

Suppose a cobalt atom in the oxidation state formed a complex with two bromide anions and four ammonia molecules. Write the chem

Katena32 [7]

Complete Question:

Suppose a cobalt atom in the +3 oxidation state formed a complex with two bromide (Br-) anions and four ammonia (NH3) molecules. write the chemical formula of this complex.

Answer:

[Co(NH₃)₄]⁺Br₂

Explanation:

The cobalt atom with +3 oxidation is represented as Co⁺³, and if it's bonded to two bromide ions, and four ammonia molecules. The molecules that are bonded to the metal atom (Co) are called complexing agents.

In the representation, we first put the molecules that surround the metal atom, forming an anion with the oxidation of the metal:

[Co(NH₃)₄]⁺³

Then, the ions are put in the formula. Because there are two bromides ion, each one with 1 minus charge, only 2 plus charged will be neutralized, and the complex will be:

[Co(NH₃)₄]⁺Br₂

4 0

3 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

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Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago