How many mols are present in 5.24 x 1024 atoms of Carbon tetra fluoride?

When magnesium chloride (MgCl2) dissolves in water, it dissociates into magnesium ions and chloride ions. How many moles of ions

timama [110]

0.448 moles of ions will form when a 42.66-gram sample of magnesium chloride is dissolved in water (option A).

<h3>How to calculate number of moles?</h3>

According to this question, when magnesium chloride (MgCl2) dissolves in water, it dissociates into magnesium ions and chloride ions as follows:

MgCl2(aq) ⟶ Mg2+(aq) + Cl−(aq)

However, if a 42.66-gram sample of magnesium chloride is dissolved in water, this means that 42.66g/molecular mass of MgCl2 will be the number of moles of the ionic products.

molecular mass of MgCl2 = 95.3g/mol

moles = 42.66g ÷ 95.3g/mol = 0.448mol

Therefore, 0.448 moles of ions will form when a 42.66-gram sample of magnesium chloride is dissolved in water.

Learn more about moles at: brainly.com/question/26416088

#SPJ1

7 0

2 years ago

A species that is isoelectronic with the nitrate ion and hence would have the same shape is [1] [2] [3] [4] [5] sulphur trioxide

Bumek [7]

Answer:

[1] sulphur trioxide

Explanation:

Isoelectronic species have the same number of valence electrons.

Valence electrons in nitrate (NO₃⁻):

5e- (N) + (3 x 6e-)(3xO) + 1e- (charge) = 24e-

Valence electrons in sulphur trioxide (SO₃):

6e- (S) + (3 x 6e-)(3xO) = 24e-

Valence electrons in sulphite (SO₃²⁻):

6e- (S) + (3 x 6e-)(3xO) + 2e- (charge) = 26e-

Valence electrons in phosphine (PH₃):

5e- (P) + (3 x 1e-)(3xH) = 8e-

Valence electrons in water (H₂O):

6e- (O) + (2 x 1e-)(2xH) = 8e-

Valence electrons in chlorite (ClO₂⁻):

7e- (Cl⁻) + (2 x 6e-)(2xO) + 1e- (charge) = 20e-

The only species isoelectronic with nitrate is sulphur trioxide. Both have trigonal planar geoemetry.

8 0

3 years ago

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

muminat

Hello!

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

We have the following data:

m (mass) = ?

n (number of moles) = 5.20 moles

MM (Molar mass of C6H12) ≈ 84.2 g/mol

Now, let's find the mass, knowing that:

$n = \dfrac{m}{MM}$

$5.20\:\:\diagup\!\!\!\!\!\!\!mol = \dfrac{m}{84.2\:g/\diagup\!\!\!\!\!\!\!mol}$

$m = 5.20*84.2$

$\boxed{\boxed{m = 437.84\:g}}\end{array}}\qquad\checkmark$

_______________________

I Hope this helps, greetings ... Dexteright02! =)

4 0

3 years ago

Read 2 more answers

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0

2 years ago

What are the mole ratios of hydrazine

juin [17]

mole ratios of hydrazine should be 1:2. I could be wrong. Are there any options to choose from?

5 0

3 years ago