Answer:
I think it is the Federal Pell Grant Program.
Explanation:
Quantum numbers<span> allow us to both simplify and dig deeper into electron configurations. Electron configurations allow us to identify energy level, subshell, and the number of electrons in those locations. If you choose to go a bit further, you can also add in x,y, or z subscripts to describe the exact orbital of those subshells (for example </span><span>2<span>px</span></span>). Simply put, electron configurations are more focused on location of electrons then anything else.
<span>
Quantum numbers allow us to dig deeper into the electron configurations by allowing us to focus on electrons' quantum nature. This includes such properties as principle energy (size) (n), magnitude of angular momentum (shape) (l), orientation in space (m), and the spinning nature of the electron. In terms of connecting quantum numbers back to electron configurations, n is related to the energy level, l is related to the subshell, m is related to the orbital, and s is due to Pauli Exclusion Principle.</span>
The answer is: " 208 g " .
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Explanation:
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The formula/ equation for density is:
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D = m / V ; That is, "mass divided by volume" ;
Density is expressed as:
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"mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
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{Note the exact equivalent: 1 cm³ = 1 mL }.
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→ The formula is: " D = m / V " ;
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in which:
"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);
"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
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We want to find the mass, "m" ; so we take the original equation/formula for the density:
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D = m / V ;
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And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
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Multiply each side of the equation by "V" ;
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V * { D = m / V } ; to get:
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V * D = m ; ↔ m = V * D ;
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Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
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m = V * D ;
m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;
→ Round to "208 g" (3 significant figures);
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The answer is: " 208 g " .
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I think the correct answer would be the third option. The criteria that could help Linda in classifying whether the gases are greenhouse gases would be gas molecules having at least one oxygen atom. Most of the greenhouse gases has an oxygen atom in their structures especially those that naturally occurs. These gases are CO2, H2O vapor and nitrous oxide.
We can use the kinematic equation
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance
we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s [if you are rounding this by significant figures]
