Answer:- 14.0 moles of hydrogen present in 2.00 moles of ![[tex]C_6H_7N](https://tex.z-dn.net/?f=%20%5Btex%5DC_6H_7N)
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Solution:- We have been given with 2.00 moles of 
and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:
= 14.0 g H
So, there are 14.0 g of hydrogen in 2.00 moles of .
Answer:
1400000 cm
Explanation:
14 km equals to 1400000 cm (14 km = 1400000 cm)
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Reactants bonds are broken and products bonds are formed
2H2(g) + O2(g) → 2H2O(1) 0 260 g 0.2068 0.180 g 2008
When 45.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 118 g. What is the percent yield? CHA(g) + 2O2(g) - CO2(g) + 2H2O(g) 73.6% 67.9% 95.2% 86.4%
For the reaction: 2503(g) + 790 kcal - 25(s) + 3O2(g), how many kcal are needed to form 1.5 moles O2(g)? 790 kcal 395 kcal 2370 kcal 411 kcal
When 3 moles of Ny are mixed with 5 moles of H2 the limiting reactant is N2(g) + 3H2(g) - 2NH3(g) H2 NH3 ОООО H20 O N₂
Answer:HNO3 + NaOH → H2O + NaNO3
Explanation:
