What us true when an ion is formed?

Which of the following can be considered evidence?

melomori [17]

Answer:

Shiny like a limousine

You're spending like a cash machine

Smile, show your golden teeth

That's how you cover up your cavities

You're glowing like a diamond ring

I saw you bought some other things

You're spending like a cash machine

To cover up your insecurities

Don't front, no needExplanation:

3455

5 0

3 years ago

A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the

Fofino [41]

Answer:

(a) $102.6g/mol$

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

$V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}$

Solving for the molarity of base we've got:

$M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L$

Now, we can compute the moles of the base as:

$n_{base}=0.425mol/L*0.1L=0.0425mol$

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

$\frac{4.36g}{0.0425mol} =102.6g/mol$

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

$102.6g/mol-16g/mol-1g/mol=85.6g/mol$

So, that atomic mass dovetails to the Rubidium's atomic mass.

Best regards.

8 0

3 years ago

21. What are the two main ways of working with clay?

Ludmilka [50]

Answer:

Diferentes tipos de arcilla

ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...

ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...

ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...

ARCILLAS “BALL CLAY” O DE BOLA. ...

CAOLIN. ...

ARCILLA REFRACTARIA. ...

BENTONITA.

Explanation:

5 0

3 years ago

Read 2 more answers

How much heat is absorbed/released when 25.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) acc

Novay_Z [31]

Answer:

429.4 kJ are absorbed in the endothermic reaction.

Explanation:

The balanced chemical equation tells us that 1168 kJ of heat are absorbed in the reaction when 4 mol of NH₃ (g) react with 5 mol O₂ (g).

So what we need is to calculates how many moles represent 25 g NH₃(g) and calculate the heat absorbed. (NH₃ is the limiting reagent)

Molar Mass NH₃ = 17.03 g/mol

mol NH₃ = 25.00 g/ 17.03 g/mol = 1.47 mol

1168 kJ /4 mol NH₃ x 1.47 mol NH₃ = 429.4 kJ

6 0

3 years ago

citric acid has an acid dissociation constant of 8.4 x 10^-4 it would be most effective for preparation of a buffer with a pH of

Sloan [31]

Answer:

(B) 3

Explanation:

Citric acid has an acid dissociation constant (Ka) of 8.4 × 10⁻⁴. When it forms a buffer with its conjugate base (citrate), we can calculate the pH using the Henderson-Hasselbalch's equation.

$pH=pKa+log\frac{[base]}{[acid]}$

The optimum range of pH is pKa ± 1. The pKa is -log Ka = -log (8.4 × 10⁻⁴) = 3.1. The buffer would be more effective for pH between 2.1 and 4.1, especially around 3.1. So the best choice is (B) 3.

5 0

3 years ago