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3 years ago

Can asteroid be a planet?

Chemistry

2 answers:

Romashka-Z-Leto [24]3 years ago

7 0

They are also known as planetoids or minor planets. I hope I helped you.

aliina [53]3 years ago

5 0

No i think they can not be

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Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol

zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = $10^{-4.8}$

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = $10^{-7.6}$

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = $10^{-1.9}$

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = $10^{1.8}$

[In-]/[HIn] = 63.10

6 0

3 years ago

What is the relationship between atoms and static electricity

Mnenie [13.5K]

They both build up to form electricity

3 0

3 years ago

Read 2 more answers

If a chemical change has taken place, which of these is most likely true?

____ [38]

<span>C. A completely new substance is formed.</span>

5 0

3 years ago

Read 2 more answers

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe

Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

$V_s=V_o-V_i=V_i\\V_o=2*V_i\\$

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

$r_o=r_i+e$

The first equation becomes

$\frac{4 \pi}{3}*r_o^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\\frac{4 \pi}{3}*(r_i+e)^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\(r_i+e)^{3} = 2 *r_i^{3} \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0$

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0

2 years ago

In the following reaction 2C6H6 + 15O2 12CO2 + 6H2O how many grams of oxygen will react with 10.47 grams of benzene (C6H6)?

IRINA_888 [86]

$2 C_{6}H_{6} + 15O_{2} ----->\ \textgreater \ 12CO_{2} + 6H_{2}O


$

$mol of benzene = \frac{mass}{Mr}$
= $\frac{10.47g}{(6 * 12) (6 * 1) g/mol}$
= 0.134 mol

mol of oxygen:
ratio of $C_{6} H_{6} : O_{2}$
= 2 : 15
= 1 : 7.5

: . mol of $O_{2}$ = 0.134mol * 7.5
= 1.01 mol

Mass of Oxygen = mol * Mr
= 1.01 mol * (16*2) g/mol
= 32.22g

Note: Mr is molar mass

8 0

3 years ago