Molar mass
CaH₂ = 42.0 g/mol
H₂ = 2.0 g/mol
<span>Balanced chemical equation :
</span>CaH2 + 2 H₂O = Ca(OH)₂ + 2 H₂
42.0 g CaH₂ --------------------- 4.0 g H2
? g CaH₂ ---------------------- 4.550 g H2
mass = 4.550 * 42.0 / 4
mass = 191,1 / 4
mass =<span> 47.775 g of CaH₂</span>
<span>hope this helps!</span>
<span>Mass Number = (Atomic Number) + (Number of Neutrons) so you solve for the Number of Neutrons and you get:
Number of Neutrons = (Mass number) - (Atomic Number)
Mass Number equals protons plus neutrons, round atomic weight to nearest whole number
Atomic Number equals number of Protons</span>
For this, you need to know 1) the mass of the hydrate and 2) the mass of the anhydrous salt. Once you have both of these, you will subtract 1) from 2) to find the mass of the water lost.
From the problem, you know that 1) = 2.000 g.
Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).
1) = 5.022 g - 3.715 g = 1.307 g
Now you can complete our original task.
Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.
Answer:
294,3
Explanation:
While g = 9,81 m/s^2 then:
![E=2[kg]*9,81[\frac{m}{s^2} ]*15[m]=294,3 [J]](https://tex.z-dn.net/?f=E%3D2%5Bkg%5D%2A9%2C81%5B%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%5D%2A15%5Bm%5D%3D294%2C3%20%5BJ%5D)
[J]
