It might refer to 3 molecules of H2 (diatomic hydrogen).
V1 = (1/3)(pi)(r12)(h1) V2 = (1/3)(pi)(xr12)(h2) V1 = V2 , which means that: (1/3)(pi)(r12)(h1) = (1/3)(pi)(xr12)(h2) Divide both sides by (1/3)(pi) and you get: (r12)(h1) = (xr12)(h2) -> (r12)(h1) = x2(r12)(h2) Divide both sides by (r12) and you get: h1 = x2(h2) -> h2 = (h1)/x2 For example: Cone1: r1 = 10, h1 = 10 Cone2: r2 = 30, h2 = (10/32) = 10/9 = 1.11111111 Then to check: Volume of a cone = (1/3)(pi)(r2)(h) V1 = (1/3)(pi)(102)(10) V1 = 1047.197551 = V2 1047.197551 = (1/3)(pi)(302)(h2) h2 = 1047.197551/((900pi)/3) h2 = 1.111111111 = 10/9
H2 +o2=h2ogive suggestion about this formulateklahaymanot tsegay
It is an expression and a term that are of equal value
-98
H2 - hydrogen.
(N2) + 3(H2) = 2(NH3)
Hydrogen peroxide
V1 = (1/3)(pi)(r12)(h1) V2 = (1/3)(pi)(xr12)(h2) V1 = V2 , which means that: (1/3)(pi)(r12)(h1) = (1/3)(pi)(xr12)(h2) Divide both sides by (1/3)(pi) and you get: (r12)(h1) = (xr12)(h2) -> (r12)(h1) = x2(r12)(h2) Divide both sides by (r12) and you get: h1 = x2(h2) -> h2 = (h1)/x2 For example: Cone1: r1 = 10, h1 = 10 Cone2: r2 = 30, h2 = (10/32) = 10/9 = 1.11111111 Then to check: Volume of a cone = (1/3)(pi)(r2)(h) V1 = (1/3)(pi)(102)(10) V1 = 1047.197551 = V2 1047.197551 = (1/3)(pi)(302)(h2) h2 = 1047.197551/((900pi)/3) h2 = 1.111111111 = 10/9
Hydrogen peroxide
You can make a simple balance. There are (12.36 * 3) moles of H You have 2*H to form H2. So take the total from ammonia and divide by two to find the moles of H2 required.
ya . H2 SO4 is an salt .(strong electrolite) because it is a salt (i=3).
1 mole H2 = 2.016g H2 = 6.022 x 1023 molecules H210g H2 x 1mol H2/2.016g H2 x 6.022 x 1023 molecules H2/1mol H2 = 3 x 1024 molecules H2 (rounded to 1 significant figure)
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
it means how many carbons there are
H2S is the formula of Hydrogen sulfide.
The reaction between nitrogen and hydrogen to form ammonia is: N2 + 3 H2 → 2 NH3 The above is the reaction for the Haber process in the industrial synthesis of ammonia. For a given proportion of 3 N2 to 2 H2 (or in ratio terms equivalent to 4.5 N2 to 3 H2), we see that H2 is the limiting reactant. Thus according to the stoichiometry of the reaction, 2 moles of H2 will form 1.33 moles of NH3.
Three: The reaction equation is N2 + 3 H2 -> 2 NH3