Giving brainliest to correct answer..<3please give answers to both

The orbital radii of four planets in our solar system is shown in the following table. Orbital Radii Planet Orbital Radii (AU) W

nalin [4]

Answer:

w and x

Explanation:

6 0

3 years ago

How many particles are in 13.5 grams of Beryllium ?

Marina86 [1]

Explanation:

I don't know this answer what Hydrogen Helium lithium

4 0

3 years ago

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How many grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate?

Korvikt [17]

Answer: There is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.

Explanation:

Chemical equation depicting reaction between sodium phosphate and aluminum carbonate is as follows.

$Al_{2}(CO_{3})_{3} + 2Na_{3}PO_{4} \rightarrow 2AlPO_{4} + 3Na_{2}CO_{3}$

As this equation contains same number of atoms on both reactant and product side. So, this equation is a balanced equation.

According to the equation, 2 moles of sodium phosphate is giving 3 moles of sodium carbonate.

Therefore, sodium carbonate formed by 5.3 moles of sodium phosphate is as follows.

$\frac{3}{2} \times 5.3 mol\\= 7.95 mol$

As number of moles is the mass of substance divided by its molar mass. So, mass of sodium carbonate ( molar mass = 105.98 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\7.95 mol = \frac{mass}{105.98 g/mol}\\mass = 842.54 g$

Thus, we can conclude that there is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.

8 0

3 years ago

A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit

nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = 1.750x10⁻³ mol aniline

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing C₆H₅NH₃⁺. C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = 2.63x10⁻⁵ (1)

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ = [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

pH = 3.70

7 0

3 years ago

15. List the substances A-E in order from most dense to least dense based on the facts provided.

Volgvan

Answer:

So 1st it is B then D then E then a then C

3 0

3 years ago