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3 years ago

You are given a model of various phases of the moon, as shown below. You can tell right away that it is inaccurate.

Chemistry

2 answers:

erma4kov [3.2K]3 years ago

8 0

Explanation:

First of all, depending on the order of which the phases in this picture are sequenced, C and D should be flipped. the 3/4 moon appears before the full moon.

Second of all, A, which is what's supposed to be a waxing crescent moon, the illuminated side should be a bit larger than what is shown. the illuminated area should be about 1/4 of the whole area of the moon circle.

Hope this helps.

a_sh-v [17]3 years ago

3 0

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I need help with this packet

Anna11 [10]

Q1. An inorganic compound is a compound where the main constituent or substance is not that of Carbon but predominantly other elements, such as I, N etc. An organic compound is one where the main substituent or main element, the element found in much greater amounts would be Carbon.

Q2. Water is considered a very good solvent, because of its ability to dissolve well with mostly all other polar compounds, and produce ions from those ionic compounds.

A. Hydrogen atoms
B. Oxygen atom.

5 0

3 years ago

239 239 - 93Np 94Pu + c

seraphim [82]

Answer:4>34.1>56

Explanation:ez

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3 years ago

The element X has three naturally occurring isotopes. The masses (amu) and % abundances of the isotopes are given in the table b

denis-greek [22]

Answer:

220.42098 amu

Explanation:

(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu

These are weighted averages.

So, we will take mass of one and multiply by abundance percentage that is provided and add them together.

In order to calculate the average atomic mass, we have to convert the percentages of abundance to decimals. So, you get

(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu

6 0

3 years ago

A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and 32 C.

lisabon 2012 [21]

Answer:

1255.4L

Explanation:

Given parameters:

P₁ = 928kpa

T₁ = 129°C

V₁ = 569L

P₂ = 319kpa

T₂ = 32°C

Unknown:

V₂ = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

take the units to the appropriate ones;

kpa to atm, °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁ = 928kpa gives 9.16atm

T₂ = 32°C gives 273 + 32 = 305K

T₁ = 129°C gives 129 + 273 = 402K

Input the values in the equation and solve for V₂;

$\frac{9.16 x 569}{402} = \frac{3.15 x V_{2} }{305}$

V₂ = 1255.4L

4 0

3 years ago

A gas sample is held at constant temperature. What happens to the pressure of the sample if the volume is doubled and the number

Iteru [2.4K]

Answer:

If the volume is doubled and the number of molecules is doubled, pressure is unchanged

Explanation:

Step 1: Data given

Temperature = constant

Volume will be doubled

Number of molecules will be doubles

Step 2:

p*V = n*R*T

⇒ gas constant and temperature are constant

Initial pressure = n*R*T / V

Initial pressure = 2*R*T/2

Initial pressure = RT

Final pressure = 4*RT / 4

Final pressure = R*T

If the volume is doubled and the number of molecules is doubled, pressure is unchanged

3 0

3 years ago