Answer:
247 ml
Explanation:
How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml
0.000150 x ? = 0.0370 moles HF
? = 0.0370/0.000150 = 247 ml
check
247 ml = 247/1000 L = 0.247
(0.247) x (0.150) =0.370 check
You calculate the mass using dimensional analysis.
We are given with
The weight of the chain per unit length which is 2.16 kg/m
We are also given with the needed length of the chain which is 7.0 m
Therefore, the mass of the chain is
7.0 m x 2.16 kg / m = mass of the chain in kg<span />
Answer:
Depending upon the mass gathered, the next process formation varies:
Nuclear fusion can kick in leading to formation of star. The nuclei fuse to together and energy is liberated in the form of light and heat.
If sufficient mass is not gathered to start nuclear fusion reaction, gaseous planet forms like Jupiter.
In third case, even though sufficient mass is present (twice the mass of Jupiter) still no fusion reaction starts. Such bodies are known as failed stars or brown dwarfs.
Answer:
C
Explanation: the clumsy definition of the mole obscures its utility. It is nearly analogous to defining a dozen as the mass of a substance that contains the same number of fundamental units as are contained in 733 g of Grade A large eggs. This definition completely obscures the utility of the dozen: that it is 12 things! Similarly, a mole is NA things. The mole is the same kind of unit as the dozen -- a certain number of things. But it differs from the dozen in a couple of ways. First, the number of things in a mole is so huge that we cannot identify with it in the way that we can identify with 12 things. Second, 12 is an important number in the English system of weights and measures, so the definition of a dozen as 12 things makes sense. However, the choice of the unusual number, 6.022 x 1023, as the number of things in a mole seems odd. Why is this number chosen? Would it not make more sense to define a mole as 1.0 x 1023 things, a nice (albeit large) integer that everyone can easily remember? To understand why the particular number, 6.022 x 1023 is used, it is necessary to resurrect an older, in some ways more sensible and useful, definition of the mole, which is grounded in the atomic weight scale addressed above.
The atomic weight scale defines the masses of atoms relative to the mass of an atom of 12C, which is assigned a mass of exactly 12.000 atomic mass units (amu). The number 12 is chosen so that the least massive atom, hydrogen, has a mass of about 1 (actually 1.008) on the scale. The atomic mass unit is a very tiny unit of mass appropriate to the scale of single atoms. Originally, of course, chemists had no idea of its value in laboratory-sized units like the gram. The early versions of the atomic weight scale were established by scientists who had no knowledge of the electron, proton, or neutron. When these were discovered in the late 19th and early 20th centuries, it turned out that the mass of an atom on the atomic weight scale was very nearly the same as the number of protons in its nucleus. This is a very useful correpondence, but it was discovered only after the weight scale had been in use for a long time.
In their desire to be able to count atoms by weighing, chemists gradually developed the concept of the "gram-atomic weight", which was defined in exact correspondence with the atomic weight scale:
1 atom of 12C weighs 12.000 amu
1 gram-atomic weight of 12C weighs 12.000 g
Answer:
a. 1810mL
Explanation:
When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:
where the temperatures must be measured in Kelvin
To convert from Celsius to Kelvin, add 273, or use the equation: 
For this problem, one must also recall that standard temperature is 0°C (or 273K).
So, ![T_1 = 273[K]](https://tex.z-dn.net/?f=T_1%20%3D%20273%5BK%5D)
, and ![T_2 = (49.4+273)[K]=322.4[K]](https://tex.z-dn.net/?f=T_2%20%3D%20%2849.4%2B273%29%5BK%5D%3D322.4%5BK%5D)
.
![\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5D%29%7D%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%29%7D)
![\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%29%7D%28322.4%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%20%29%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29%7D%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29)
![1810.04571428[mL]=V_2](https://tex.z-dn.net/?f=1810.04571428%5BmL%5D%3DV_2)
Adjusting for significant figures, this gives ![V_2=1810[mL]](https://tex.z-dn.net/?f=V_2%3D1810%5BmL%5D)
