<span>True
</span><span>True
</span><span>False*
</span><span>False*
</span><span>True
</span><span>True
</span><span>False
A,B,AB,O
10.)?
11.)</span><span>water
carbon dioxide
12.)</span><span>geocentric
</span>13.)<span>Juptier</span>
This question involves the concepts of density, volume, and mass.
The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".
<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>
It is given that:
Mass of one mole = 24 grams
Mass of 6 x 10²³ atoms = 24 grams
Mass of 1 atom = 
= 4 x 10⁻²³ grams
<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>
where,
= density = 1.7 grams/cm³- m = mass of single atom = 4 x 10⁻²³ grams
- V = volume of single atom = ?
Therefore,
V = 2.35 x 10⁻²³ cm³
<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>
The atom is in a spherical shape. Hence, its Volume can be given as follows:
![V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}](https://tex.z-dn.net/?f=V%20%3D%5Cfrac%7B%5Cpi%20d%5E3%7D%7B6%7D%5C%5C%5C%5Cd%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6V%7D%7B%5Cpi%7D%7D%5C%5C%5C%5Cd%3D%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6%282.35%5C%20x%5C%2010%5E%7B-23%7D%5C%20cm%5E3%29%7D%7B%5Cpi%7D%7D)
d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m
Learn more about density here:
brainly.com/question/952755
Answer:
sun, jupiter, earth, moon
Explanation:
how big they are
Answer:
a) Temperatura, b) Temperature, c) Constant
, d) None of these
, e) Gibbs enthalpy and free energy (G)
Explanation:
a) the expression for ideal gases is PV = nRT
Temperature
b) The internal energy is E = K T
Temperature
c) S = ΔQ/T
In an isolated system ΔQ is zero, entropy is constant
Constant
d) all parameters change when changing status
None of these
e) Gibbs enthalpy and free energy
The normal stress follows the formula written below:
σ = F/A
There are two types of stress, axial and tangential. Since we are only given with the dimension of the radius (and not the length), the possible stress is axial. So, the area is,
A = πr² = π(0.75 in)² = 1.767 in²
So,
σ = F/A = 500 lb/1.767 in² = <em>282.94 psi</em>
