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 .
Im not positive but i think it is When it is a gas?
<u>0.219 moles </u><u>moles are present in the flask when the </u><u>pressure </u><u>is 1.10 atm and the temperature is 33˚c.</u>
What is ideal gas constant ?
- The ideal gas constant is calculated to be 8.314J/K⋅ mol when the pressure is in kPa.
- The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
- The combined gas law relates pressure, volume, and temperature of a gas.
We simple use this formula-
The basic formula is PV = nRT where. P = Pressure in atmospheres (atm) V = Volume in Liters (L) n = of moles (mol) R = the Ideal Gas Law Constant.
68F = 298.15K
V = nRT/P = 0.2 * 0.08206 * 298.15K / (745/760) = 4.992Liters
n = PV/RT = 1.1atm*4.992L/(0.08206Latm/molK * 306K)
n = 0.219 moles
Therefore, 0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.
Learn more about ideal gas constant
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Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>
