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3 years ago

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:

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 closest to the Sun. 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 natural satellite of the planet Earth. It is the second most brightest object in the Earth's sky after Sun.

Since Mercury is much closer to the Sun than the Moon. Therefore, Mercury is much hotter than the Moon.

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