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ladessa [460]
3 years ago
7

2. What was the primary goal of the Apollo program?

Chemistry
2 answers:
Ivenika [448]3 years ago
7 0
The right answer is bbbbb
babunello [35]3 years ago
3 0

Answer:

B

Explanation:

to conduct scientific exploration of the moon

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Question 6(Multiple Choice Worth 2 points)
Rina8888 [55]
4.2 × 10²² atoms Al × (1 mol Al / 6.022 × 10²³) = moles Al

Last one: fraction 1 mole Al over 6.022 × 10²³ atoms Al
8 0
3 years ago
A sample of an unknown metal has a mass of 120.4 g. As the sample cools from 90.5°C to 25.7°C, it releases 7020 J of energy. Usi
Dafna1 [17]

Answer:

Explanation:

sheesh i wouldn’t know

6 0
3 years ago
What is the main reason Mercury is much hotter than the Moon?
Vika [28.1K]

Answer:

The correct option is: c. Mercury is closer to the Sun                                

Explanation:

Mercury is the first and the smallest planet of the Solar system that is present <u>closest to the Sun</u>. The surface temperature of this planet ranges from -173 °C-427 °C or 100-700 K.

Due to the small size and high surface temperature of the planet, the gravity can not retain a significant atmosphere.

Whereas, Moon is the only permanent <u>natural satellite of the planet Earth.</u> It is the second most brightest object in the Earth's sky after Sun.

<u>Since Mercury is much closer to the Sun than the Moon. Therefore, Mercury is much hotter than the Moon.</u>

6 0
3 years ago
The absorbance of a 0.45 mol dm–3 solution of an aromatic amino acid, 3.0 cm thick is 0.22 at a wavelength of 295 nm:
34kurt

Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times C\times l

where,

A = absorbance of solution = 0.22

C = concentration of solution = 0.45 mol/dm^3=0.45 mol/L=0.45 M

1 dm^3 = 1 L

l = length of the cell = 3.0 cm

\epsilon = molar absorptivity of this solution = ?

Now put all the given values in the above formula, we get the molar absorptivity of this solution.

0.22=\epsilon \times (0.45 M)\times (3.0 cm)

\epsilon=0.163 M^{-1}cm^{-1}

Therefore, the molar absorptivity of this solution is, 1.93\times 10^{4}M^{-1}cm^{-1}

b) A=\log \frac{I_o}{I_t}

T=\frac{I_t}{I_o}

A=\log \frac{1}{T}

A = 2 × 0.22 =0.44

I_o,I_t = Intensities of Incident light and transmitted light respectively

T = Transmittance

0.44=\log \frac{1}{T}

T = 0.3630

c) I_o=x

I_t=65\% of x=0.65 x

Thickness of cell = l' =?

c = 0.75 mol/ dm^3=0.75 mol/L=0.75 M

A=\log \frac{I_o}{I_t}=\epsilon \times C\times l

\log \frac{x}{0.65x}=0.163 M^{-1}cm^{-1}\times 0.45 M\times l'

l' = 1.53 cm

d) No, we cannot calculate the absorbance at 590 nm from the given data. This is because absorbance at this wavelength  can be observe experimentally.

8 0
4 years ago
How many seconds does it take to deposit 0.94 g of Ni on a decorative drawer handle when 14.9 A is passed through a Ni(NO3)2 sol
Goshia [24]

Answer:

Time take to deposit Ni is 259.02 sec.

Explanation:

Given:

Current I = 14.9 A

Faraday constant = 96485 \frac{C}{mole}

Molar mass of Ni = 58.69 \frac{g}{mole}

Mass of Ni = 0.94 g

First find the no. moles in Ni solution,

Moles of Ni = \frac{0.94}{58.69}

                   = 0.02 mol

From the below reaction,

  Ni^{2+}  + 2e ⇆ Ni_{(s)}

Above reaction shows "1 mol of Ni^{2+} requires 2 mol of electron to form 1 mol of Ni_{(s)} "

So for finding charge flow in this reaction we write,

    = 0.02 \times \frac{2  }{1 }  \times 96485 \frac{C}{mol}

Charge flow = 3859.4 C

For finding time of reaction,

  I = \frac{q}{t}

Where q = charge flow

   t = \frac{q}{I}

   t = \frac{3859.4}{14.9}

   t = 259.02 sec

Therefore, time take to deposit Ni is 259.02 sec.

3 0
3 years ago
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