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kramer
3 years ago
10

What is the name of the visible surface of the Sun? (And surface doesn’t imply solid, as this surface’s density is 6000 times lo

wer than the density of air)
a. Photosphere

b. Chromosphere

c. Corona

d. Core

e.Radiative Zone
Chemistry
1 answer:
Fiesta28 [93]3 years ago
4 0

Answer:

The correct option is: a. Photosphere

Explanation:

Sun is the brightest star in the sky of our planet Earth and the principle component of the Solar System. It is a <u>gaseous object</u> that is made up of hot plasma and contributes 99.86% to the total mass of our Solar System.

<u>The Sun does not have a clearly defined surface.</u> The <u>visible surface of the Sun is known as the Photosphere</u>, which is the deepest visible portion of the Sun.

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Instead of using ratios for back titrations we can also use molarities, if our solutions are standardized. A 0.196 g sample of a
kvv77 [185]

Answer:

The mass percent of Al(OH)₃ is 15.3%

Explanation:

The reaction is:

Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

The excess acid is neutralized with a solution of sodium hidroxide, in the reaction:

NaOH + HCl = NaCl + H₂O

The total moles of HCl is:

n_{HCl,total} =M_{HCl} *V_{HCl} =0.111*0.025=2.78x10^{-3} moles

From the second titration, the moles of excess of HCl is:

n_{HCl,excess} =n_{NaOH} =M_{NaOH} *V_{NaOH} =0.132*0.01105=1.46x10^{-3} moles

The difference between the total and excess of HCl, it can be know the moles that reacts with the aluminum hydroxide, is:

n_{HCl,reacts} =n_{HCl,total}-n_{HCl,excess} =2.78x10^{-3} moles-1.46x10^{-3} moles=1.32x10^{-3} moles

The ratio between HCl and Al(OH)₃ is 3:1. The MW for aluminum hydroxide is 78 g/mol, thus:

m_{Al(OH)3} =1.32x10^{-3} molesHCl*\frac{1molAl(OH)3}{3molesHCl} *\frac{78gAl(OH)3}{1molAl(OH)3} =0.03g

The percentage of Al(OH)₃ is:

Percentage-Al(OH)3=\frac{m_{Al(OH)3} }{m_{antiacid} } *100=\frac{0.03}{0.196} =15.3%

3 0
3 years ago
The answer to the problem: 3.542 - 2.1 = _____, written to the correct number of significant digits is:
VMariaS [17]

1.442 or 1.4

Round it to the nearest number

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3 years ago
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Which of the following processes would you predict to have a positive value for ΔSºreaction ?
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5 0
3 years ago
A sample of pure NO2NO2 is heated to 335 ∘C∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2N
Alborosie

Answer:

k_c = 1. 1 × 10⁻²

Explanation:

Given that:

Temperature = 335 ° C = (335+ 273)K = 608

Pressure = 0.750 atm

Volume = 1 Litre

number of moles of NO2 = ???

Rate Constant =0.0821 L atm /K/mol

Using the Ideal gas equation

PV = nRT

n = \frac{PV}{RT}

n = \frac{0.75*1}{0.0821*608}

n = 0.015

n = 1.5 × 10⁻² mole

Density = 0.525 g/L

The equation for the reaction can be illustrated as:

                     2NO2(g)         ⇌          2NO(g)         +         O2(g)

For the ICE table; we have:

 

Initial                 x                                   0                              0

Change            -2y                               + 2y                          +y

Equilibrium        (x - 2y)                        2y                             y

Total moles at equilibrium = (x-2y)+2y+y

= x + y moles

However,

1.5 × 10⁻² mole of the mixture has a mass of 0.525 g

i.e x + y moles = 1.5 × 10⁻² mole

Now, molar mass of 1 mole of NO2 = 46g/mol

Since number of moles = \frac{mass}{molar mass}

mass of (x-2y) moles = 46 × (x-2y) g

Molar mass of NO = 30 g/mol

Also, mass of NO = 2y × 30 = 60y

Molar mass of O2 = 32 g/mol

Mass of O2 = y × 32 = 32y

Total mass = ( 46x - 90y)+60y+32y = 0.525

46x = 0.525

x = \frac{0.525}{46}

x = 0.0114

x = 1.14 × 10⁻²

x + y moles = 1.5 × 10⁻²

y =  1.5 × 10⁻² -  1.14 × 10⁻²

y = 0.0036

y = 3.6 × 10⁻³

At equilibrium

[NO2] = ( 1.14 - 2(0.36))× 10⁻² = 4.2 × 10⁻³ M

[NO] = 2 ( 3.6 × 10⁻³)  = 7.2 × 10⁻³ M

[O2] = 3.6 × 10⁻³ M

k_c = \frac{[NO]^2[O_2]}{[NO_2]^2}

k_c = \frac{(7.2*10^{-3})^2(3.6*10^{-3})}{(4.2*10^-3)^2}

k_c = 0.011

k_c = 1. 1 × 10⁻²

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3 years ago
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